RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.4

Exercise 6.4

Page-6.21

Question 1:

Find the following product:
2a³(3a + 5b)

Answer 1:

To find the product, we will use distributive law as follows:

$$\begin{gathered} 2a^3\left(3a+5b\right) \\ =2a^3\times 3a+2a^3\times 5b \\ =\left(2\times 3\right)\left(a^3\times a\right)+\left(2\times 5\right)a^{3}b \\ =\left(2\times 3\right)a^{3+1}+\left(2\times 5\right)a^{3}b \\ =6a^4+10a^{3}b \end{gathered}$$

Thus, the answer is $6a^4+10a^{3}b$.

Question 2:

Find the following product:
−11a(3a + 2b)

Answer 2:

To find the product, we will use distributive law as follows:

$$\begin{gathered} -11a\left(3a+2b\right) \\ =\left(-11a\right)\times 3a+\left(-11a\right)\times 2b \\ =\left(-11\times 3\right)\times \left(a\times a\right)+\left(-11\times 2\right)\times \left(a\times b\right) \\ =\left(-33\right)\times \left(a^{1+1}\right)+\left(-22\right)\times \left(a\times b\right) \\ =-33a^2-22ab \end{gathered}$$

Thus, the answer is $-33a^2-22ab$.

Question 3:

Find the following product:
−5a(7a − 2b)

Answer 3:

To find the product, we will use distributive law as follows:

$$\begin{gathered} -5a\left(7a-2b\right) \\ =\left(-5a\right)\times 7a+\left(-5a\right)\times \left(-2b\right) \\ =\left(-5\times 7\right)\times \left(a\times a\right)+\left(-5\times \left(-2\right)\right)\times \left(a\times b\right) \\ =\left(-35\right)\times \left(a^{1+1}\right)+\left(10\right)\times \left(a\times b\right) \\ =-35a^2+10ab \end{gathered}$$

Thus, the answer is $-35a^2+10ab$.

Question 4:

Find the following product:
−11y²(3y + 7)

Answer 4:

To find the product, we will use distributive law as follows:

$$\begin{gathered} -11y^2\left(3y+7\right) \\ =\left(-11y^2\right)\times 3y+\left(-11y^2\right)\times 7 \\ =\left(-11\times 3\right)\left(y^2\times y\right)+\left(-11\times 7\right)\times \left(y^2\right) \\ =\left(-33\right)\left(y^{2+1}\right)+\left(-77\right)\times \left(y^2\right) \\ =-33y^3-77y^2 \end{gathered}$$

Thus, the answer is $-33y^3-77y^2$.

Question 5:

Find the following product:
$\frac{6x}{5}(x^3+y^3)$

Answer 5:

To find the product, we will use distributive law as follows:

$$\begin{gathered} \frac{6x}{5}\left(x^3+y^3\right) \\ =\frac{6x}{5}\times x^3+\frac{6x}{5}\times y^3 \\ =\frac{6}{5}\times \left(x\times x^3\right)+\frac{6}{5}\times \left(x\times y^3\right) \\ =\frac{6}{5}\times \left(x^{1+3}\right)+\frac{6}{5}\times \left(x\times y^3\right) \\ =\frac{6x^4}{5}+\frac{6x{y}^3}{5} \end{gathered}$$

Thus, the answer is $\frac{6x^4}{5}+\frac{6x{y}^3}{5}$.

Question 6:

xy(x³ − y³)

Answer 6:

To find the product, we will use the distributive law in the following way:

$$\begin{gathered} xy\left(x^3-y^3\right) \\ =xy\times x^3-xy\times y^3 \\ =\left(x\times x^3\right)\times y-x\times \left(y\times y^3\right) \\ =x^{1+3}y-x{y}^{1+3} \\ =x^{4}y-x{y}^4 \end{gathered}$$

Thus, the answer is $x^{4}y-x{y}^4$.

Question 7:

Find the following product:
0.1y(0.1x⁵ + 0.1y)

Answer 7:

To find the product, we will use distributive law as follows:

$$\begin{gathered} 0.1y\left(0.1x^5+0.1y\right) \\ =\left(0.1y\right)\left(0.1x^5\right)+\left(0.1y\right)\left(0.1y\right) \\ =\left(0.1\times 0.1\right)\left(y\times x^5\right)+\left(0.1\times 0.1\right)\left(y\times y\right) \\ =\left(0.1\times 0.1\right)\left(x^5\times y\right)+\left(0.1\times 0.1\right)\left(y^{1+1}\right) \\ =0.01x^{5}y+0.01y^2 \end{gathered}$$

Thus, the answer is $0.01x^{5}y+0.01y^2$.

Question 8:

Find the following product:
$\left(-\frac{7}{4}a{b}^{2}c-\frac{6}{25}{a}^2{c}^2\right)(-50a^2{b}^2{c}^2)$

Answer 8:

To find the product, we will use distributive law as follows:

$$\begin{gathered} \left(-\frac{7}{4}a{b}^{2}c-\frac{6}{25}{a}^2{c}^2\right)\left(-50a^2{b}^2{c}^2\right) \\ =\left\{\left(-\frac{7}{4}a{b}^{2}c\right)\left(-50a^2{b}^2{c}^2\right)\right\}-\left\{\left(\frac{6}{25}{a}^2{c}^2\right)\left(-50a^2{b}^2{c}^2\right)\right\} \\ =\left\{\left\{-\frac{7}{4}\times \left(-50\right)\right\}\left(a\times a^2\right)\times \left(b^2\times b^2\right)\times \left(c\times c^2\right)\right\}-\left\{\left(\frac{6}{25}\right)\left(-50\right)\left(a^2\times a^2\right)\times \left(b^2\right)\times \left(c^2\times c^2\right)\right\} \\ =\left\{-\frac{7}{4}\times \left(-50\right)\right\}\left(a^{1+2}{b}^{2+2}{c}^{1+2}\right)-\left\{\left(\frac{6}{25}\right)\left(-50\right)\left(a^{2+2}{b}^2{c}^{2+2}\right)\right\} \\ =\frac{175}{2}{a}^3{b}^4{c}^3-\left(-12a^4{b}^2{c}^4\right) \\ =\frac{175}{2}{a}^3{b}^4{c}^3+12a^4{b}^2{c}^4 \end{gathered}$$

Thus, the answer is $\frac{175}{2}{a}^3{b}^4{c}^3+12a^4{b}^2{c}^4$.

Question 9:

Find the following product:
$-\frac{8}{27}xyz\left(\frac{3}{2}xy{z}^2-\frac{9}{4}x{y}^2{z}^3\right)$

Answer 9:

To find the product, we will use the distributive law in the following way:

$$\begin{gathered} -\frac{8}{27}xyz\left(\frac{3}{2}xy{z}^2-\frac{9}{4}x{y}^2{z}^3\right) \\ =\left\{\left(-\frac{8}{27}xyz\right)\left(\frac{3}{2}xy{z}^2\right)\right\}-\left\{\left(-\frac{8}{27}xyz\right)\left(\frac{9}{4}x{y}^2{z}^3\right)\right\} \\ =\left\{\left(-\frac{8}{27}\times \frac{3}{2}\right)\left(x\times x\right)\times \left(y\times y\right)\times \left(z\times z^2\right)\right\}-\left\{\left(-\frac{8}{27}\times \frac{9}{4}\right)\left(x\times x\right)\times \left(y\times y^2\right)\times \left(z\times z^3\right)\right\} \\ =\left\{\left(-\frac{8}{27}\times \frac{3}{2}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}-\left\{\left(-\frac{8}{27}\times \frac{9}{4}\right)\left(x^{1+1}{y}^{1+2}{z}^{1+3}\right)\right\} \\ =\left\{\left(-\frac{{\cancel{8}}^4}{{\cancel{27}}_9}\times \frac{\cancel{3}}{\cancel{2}}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}-\left\{\left(-\frac{{\cancel{8}}^2}{{\cancel{27}}_3}\times \frac{\cancel{9}}{\cancel{4}}\right)\left(x^{1+1}{y}^{1+2}{z}^{1+3}\right)\right\} \\ =-\frac{4}{9}{x}^2{y}^2{z}^3+\frac{2}{3}{x}^2{y}^3{z}^4 \end{gathered}$$

Thus, the answer is $-\frac{4}{9}{x}^2{y}^2{z}^3+\frac{2}{3}{x}^2{y}^3{z}^4$.

Question 10:

Find the following product:
$-\frac{4}{27}xyz\left(\frac{9}{2}{x}^{2}yz-\frac{3}{4}xy{z}^2\right)$

Answer 10:

To find the product, we will use distributive law as follows:

$$\begin{gathered} -\frac{4}{27}xyz\left(\frac{9}{2}{x}^{2}yz-\frac{3}{4}xy{z}^2\right) \\ =\left\{\left(-\frac{4}{27}xyz\right)\left(\frac{9}{2}{x}^{2}yz\right)\right\}-\left\{\left(-\frac{4}{27}xyz\right)\left(\frac{3}{4}xy{z}^2\right)\right\} \\ =\left\{\left(-\frac{4}{27}\times \frac{9}{2}\right)\left(x^{1+2}{y}^{1+1}{z}^{1+1}\right)\right\}-\left\{\left(-\frac{4}{27}\times \frac{3}{4}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\} \\ =\left\{\left(-\frac{{\cancel{4}}^2}{{\cancel{27}}_3}\times \frac{\cancel{9}}{\cancel{2}}\right)\left(x^{1+2}{y}^{1+1}{z}^{1+1}\right)\right\}-\left\{\left(-\frac{{\cancel{4}}^1}{{\cancel{27}}_9}\times \frac{\cancel{3}}{\cancel{4}}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\} \\ =-\frac{2}{3}{x}^3{y}^2{z}^2+\frac{1}{9}{x}^2{y}^2{z}^3 \end{gathered}$$

Thus, the answer is $-\frac{2}{3}{x}^3{y}^2{z}^2+\frac{1}{9}{x}^2{y}^2{z}^3$.

Question 11:

Find the following product:
1.5x(10x²y − 100xy²)

Answer 11:

To find the product, we will use distributive law as follows:

$$\begin{gathered} 1.5x\left(10x^{2}y-100x{y}^2\right) \\ =\left(1.5x\times 10x^{2}y\right)-\left(1.5x\times 100x{y}^2\right) \\ =\left(15x^{1+2}y\right)-\left(150x^{1+1}{y}^2\right) \\ =15x^{3}y-150x^2{y}^2 \end{gathered}$$

Thus, the answer is $15x^{3}y-150x^2{y}^2$.

Question 12:

Find the following product:
4.1xy(1.1x − y)

Answer 12:

To find the product, we will use distributive law as follows:

$$\begin{gathered} 4.1xy\left(1.1x-y\right) \\ =\left(4.1xy\times 1.1x\right)-\left(4.1xy\times y\right) \\ =\left\{\left(4.1\times 1.1\right)\times xy\times x\right\}-\left(4.1xy\times y\right) \\ =\left(4.51x^{1+1}y\right)-\left(4.1x{y}^{1+1}\right) \\ =4.51x^{2}y-4.1x{y}^2 \end{gathered}$$

Thus, the answer is $4.51x^{2}y-4.1x{y}^2$.

Question 13:

Find the following product:
250.5xy$\left(xz+\frac{y}{10}\right)$

Answer 13:

To find the product, we will use distributive law as follows:

$$\begin{gathered} 250.5xy\left(xz+\frac{y}{10}\right) \\ =250.5xy\times xz+250.5xy\times \frac{y}{10} \\ =250.5x^{1+1}yz+25.05x{y}^{1+1} \\ =250.5x^{2}yz+25.05x{y}^2 \end{gathered}$$

Thus, the answer is $250.5x^{2}yz+25.05x{y}^2$.

Question 14:

Find the following product:
$\frac{7}{5}{x}^{2}y\left(\frac{3}{5}x{y}^2+\frac{2}{5}x\right)$

Answer 14:

To find the product, we will use distributive law as follows:

$$\begin{gathered} \frac{7}{5}{x}^{2}y\left(\frac{3}{5}x{y}^2+\frac{2}{5}x\right) \\ =\frac{7}{5}{x}^{2}y\times \frac{3}{5}x{y}^2+\frac{7}{5}{x}^{2}y\times \frac{2}{5}x \\ =\frac{21}{25}{x}^{2+1}{y}^{1+2}+\frac{14}{25}{x}^{2+1}y \\ =\frac{21}{25}{x}^3{y}^3+\frac{14}{25}{x}^{3}y \end{gathered}$$

Thus, the answer is $\frac{21}{25}{x}^3{y}^3+\frac{14}{25}{x}^{3}y$.

Question 15:

Find the following product:
$\frac{4}{3}a(a^2 + b^2 - 3c^2)$

Answer 15:

To find the product, we will use distributive law as follows:

$$\begin{gathered} \frac{4}{3}a\left(a^2+b^2-3c^2\right) \\ =\frac{4}{3}a\times a^2+\frac{4}{3}a\times b^2-\frac{4}{3}a\times 3c^2 \\ =\frac{4}{3}{a}^{1+2}+\frac{4}{3}a{b}^2-4a{c}^2 \\ =\frac{4}{3}{a}^3+\frac{4}{3}a{b}^2-4a{c}^2 \end{gathered}$$

Thus, the answer is $\frac{4}{3}{a}^3+\frac{4}{3}a{b}^2-4a{c}^2$.

Question 16:

Find the product 24x² (1 − 2x) and evaluate its value for x = 3.

Answer 16:

To find the product, we will use distributive law as follows:

$$\begin{gathered} 24x^2\left(1-2x\right) \\ =24x^2\times 1-24x^2\times 2x \\ =24x^2-48x^{1+2} \\ =24x^2-48x^3 \end{gathered}$$

Substituting  x = 3 in the result, we get:

$$\begin{gathered} 24x^2-48x^3 \\ =24{\left(3\right)}^2-48{\left(3\right)}^3 \\ =24\times 9-48\times 27 \\ =216-1296 \\ =-1080 \end{gathered}$$

Thus, the product is $(24x^2-48x^3) \mathrm{and} \mathrm{its} \mathrm{value} \mathrm{fo}r x = 3 \mathrm{is} (-1080)$.

Question 17:

Find the product −3y(xy + y²) and find its value for x = 4 and y = 5.

Answer 17:

To find the product, we will use distributive law as follows:​

$$\begin{gathered} -3y\left(xy+y^2\right) \\ =-3y\times xy+\left(-3y\right)\times y^2 \\ =-3x{y}^{1+1}-3y^{1+2} \\ =-3x{y}^2-3y^3 \end{gathered}$$

Substituting x = 4 and y = 5 in the result, we get:

$$\begin{gathered} -3x{y}^2-3y^3 \\ =-3\left(4\right){\left(5\right)}^2-3{\left(5\right)}^3 \\ =-3\left(4\right)\left(25\right)-3\left(125\right) \\ =-300-375 \\ =-675 \end{gathered}$$

Thus, the product is ($-3x{y}^2-3y^3$), and its value for ​x = 4 and y = 5 is ($-$675).

Question 18:

Multiply $-\frac{3}{2}{x}^2{y}^3 \mathrm{by} (2x - y)$ and verify the answer for x = 1 and y = 2.

Answer 18:

To find the product, we will use distributive law as follows:

$$\begin{gathered} -\frac{3}{2}{x}^2{y}^3\times \left(2x-y\right) \\ =\left(-\frac{3}{2}{x}^2{y}^3\times 2x\right)-\left(-\frac{3}{2}{x}^2{y}^3\times y\right) \\ =\left(-3x^{2+1}{y}^3\right)-\left(-\frac{3}{2}{x}^2{y}^{3+1}\right) \\ =-3x^3{y}^3+\frac{3}{2}{x}^2{y}^4 \end{gathered}$$

Substituting x = 1 and y = 2 in the result, we get:

$$\begin{gathered} -3x^3{y}^3+\frac{3}{2}{x}^2{y}^4 \\ =-3{\left(1\right)}^3{\left(2\right)}^3+\frac{3}{2}{\left(1\right)}^2{\left(2\right)}^4 \\ =-3\times 1\times 8+\frac{3}{2}\times 1\times 16 \\ =-24+24 \\ =0 \end{gathered}$$

Thus, the product is $-3x^3{y}^3+\frac{3}{2}{x}^2{y}^4$, and its value for ​x = 1 and y = 2 is 0.

Question 19:

Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y²(2 − 3x)
(ii) −3x(y² + z²)
(iii) z²(x − y)
(iv) xz(x² + y²)

Answer 19:

(i) To find the product, we will use distributive law as follows:

$$\begin{gathered} 15y^2\left(2-3x\right) \\ =15y^2\times 2-15y^2\times 3x \\ =30y^2-45x{y}^2 \end{gathered}$$

Substituting x = $-$1 and y = 0.25 in the result, we get:

$$\begin{gathered} 30y^2-45x{y}^2 \\ =30{\left(0.25\right)}^2-45\left(-1\right){\left(0.25\right)}^2 \\ =30\times 0.0625-\left\{45\times \left(-1\right)\times 0.0625\right\} \\ =30\times 0.0625-\left\{45\times \left(-1\right)\times 0.0625\right\} \\ =1.875-\left(-2.8125\right) \\ =1.875+2.8125 \\ =4.6875 \end{gathered}$$

(ii) To find the product, we will use distributive law as follows:

$$\begin{gathered} -3x\left(y^2+z^2\right) \\ =-3x\times y^2+\left(-3x\right)\times z^2 \\ =-3x{y}^2-3x{z}^2 \end{gathered}$$

Substituting x = $-$1, y = 0.25​ and z = 0.05​ in the result, we get:

$$\begin{gathered} -3x{y}^2-3x{z}^2 \\ =-3\left(-1\right){\left(0.25\right)}^2-3\left(-1\right){\left(0.05\right)}^2 \\ =-3\left(-1\right)\left(0.0625\right)-3\left(-1\right)\left(0.0025\right) \\ =01875+0.0075 \\ =0.195 \end{gathered}$$

(iii) To find the product, we will use distributive law as follows:

$$\begin{gathered} z^2\left(x-y\right) \\ =z^2\times x-z^2\times y \\ =x{z}^2-y{z}^2 \end{gathered}$$

Substituting x = $-$1, y = 0.25​ and z = 0.05​ in the result, we get:​

$$\begin{gathered} x{z}^2-y{z}^2 \\ =\left(-1\right){\left(0.05\right)}^2-\left(0.25\right){\left(0.05\right)}^2 \\ =\left(-1\right)\left(0.0025\right)-\left(0.25\right)\left(0.0025\right) \\ =-0.0025-0.000625 \\ =-0.003125 \end{gathered}$$

(iv) To find the product, we will use distributive law as follows:

$$\begin{gathered} xz\left(x^2+y^2\right) \\ =xz\times x^2+xz\times y^2 \\ =x^{3}z+x{y}^{2}z \end{gathered}$$

Substituting x = $-$1, y = 0.25​ and z = 0.05​ in the result, we get:​

$$\begin{gathered} x^{3}z+x{y}^{2}z \\ ={\left(-1\right)}^3\left(0.05\right)+\left(-1\right){\left(0.25\right)}^2\left(0.05\right) \\ =\left(-1\right)\left(0.05\right)+\left(-1\right)\left(0.0625\right)\left(0.05\right) \\ =-0.05-0.003125 \\ =-0.053125 \end{gathered}$$

Question 20:

Simplify:
(i) 2x²(x³ − x) − 3x(x⁴ + 2x) − 2(x⁴ − 3x²)
(ii) x³y(x² − 2x) + 2xy(x³ − x⁴)
(iii) 3a² + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x² − 1) + 4x² + 4
(v) a(b − c) − b(c − a) − c(a − b)
(vi) a(b − c) + b(c − a) + c(a − b)
(vii) 4ab(a − b) − 6a²(b − b²) − 3b²(2a² − a) + 2ab(b − a)
(viii) x²(x² + 1) − x³(x + 1) − x(x³ − x)
(ix) 2a² + 3a(1 − 2a³) + a(a + 1)
(x) a²(2a − 1) + 3a + a³ − 8
(xi) $\frac{3}{2}{x}^2(x^2-1)+\frac{1}{4}{x}^2(x^2+x)-\frac{3}{4}x(x^3-1)$
(xii) a²b(a − b²⁾ + ab²(4ab − 2a²) − a³b(1 − 2b)
(xiii) a²b(a³ − a + 1) − ab(a⁴ − 2a² + 2a) − b (a³ − a² − 1)

Answer 20:

(i) To simplify, we will use distributive law as follows:

$$\begin{gathered} 2x^2\left(x^3-x\right)-3x\left(x^4+2x\right)-2\left(x^4-3x^2\right) \\ =2x^5-2x^3-3x^5-6x^2-2x^4+6x^2 \\ =2x^5-3x^5-2x^4-2x^3-6x^2+6x^2 \\ =-x^5-2x^4-2x^3 \end{gathered}$$

(ii) To simplify, we will use distributive law as follows:​

$$\begin{gathered} x^{3}y\left(x^2-2x\right)+2xy\left(x^3-x^4\right) \\ =x^{5}y-2x^{4}y+2x^{4}y-2x^{5}y \\ =x^{5}y-2x^{5}y-2x^{4}y+2x^{4}y \\ =-x^{5}y \end{gathered}$$

(iii) To simplify, we will use distributive law as follows:​

$$\begin{gathered} 3a^2+2\left(a+2\right)-3a\left(2a+1\right) \\ =3a^2+2a+4-6a^2-3a \\ =3a^2-6a^2+2a-3a+4 \\ =-3a^2-a+4 \end{gathered}$$

(iv) To simplify, we will use distributive law as follows:

$$\begin{gathered} x\left(x+4\right)+3x\left(2x^2-1\right)+4x^2+4 \\ =x^2+4x+6x^3-3x+4x^2+4 \\ =x^2+4x^2+4x-3x+6x^3+4 \\ =5x^2+x+6x^3+4 \end{gathered}$$

(v) To simplify, we will use distributive law as follows:​

$$\begin{gathered} a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right) \\ =ab-ac-bc+ba-ca+cb \\ =ab+ba-ac-ca-bc+cb \\ =2ab-2ac \end{gathered}$$

(vi) To simplify, we will use distributive law as follows:​

$$\begin{gathered} a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right) \\ =ab-ac+bc-ba+ca-cb \\ =ab-ba-ac+ca+bc-cb \\ =0 \end{gathered}$$

(vii) To simplify, we will use distributive law as follows:​

$$\begin{gathered} 4ab\left(a-b\right)-6a^2\left(b-b^2\right)-3b^2\left(2a^2-a\right)+2ab\left(b-a\right) \\ =4a^{2}b-4a{b}^2-6a^{2}b+6a^2{b}^2-6b^2{a}^2+3b^{2}a+2a{b}^2-2a^{2}b \\ =4a^{2}b-6a^{2}b-2a^{2}b-4a{b}^2+3b^{2}a+2a{b}^2+6a^2{b}^2-6b^2{a}^2 \\ =-4a^{2}b+a{b}^2 \end{gathered}$$

(viii) To simplify, we will use distributive law as follows:​

$$\begin{gathered} x^2\left(x^2+1\right)-x^3\left(x+1\right)-x\left(x^3-x\right) \\ =x^4+x^2-x^4-x^3-x^4+x^2 \\ =x^4-x^4-x^4-x^3+x^2+x^2 \\ =-x^4-x^3+2x^2 \end{gathered}$$

(ix) To simplify, we will use distributive law as follows:​

$$\begin{gathered} 2a^2+3a\left(1-2a^3\right)+a\left(a+1\right) \\ =2a^2+3a-6a^4+a^2+a \\ =2a^2+a^2+3a+a-6a^4 \\ =3a^2+4a-6a^4 \end{gathered}$$

(x) To simplify, we will use distributive law as follows:​

$$\begin{gathered} a^2\left(2a-1\right)+3a+a^3-8 \\ =2a^3-a^2+3a+a^3-8 \\ =2a^3+a^3-a^2+3a-8 \\ =3a^3-a^2+3a-8 \end{gathered}$$

(xi) To simplify, we will use distributive law as follows:​

$$\begin{gathered} \frac{3}{2}{x}^2\left(x^2-1\right)+\frac{1}{4}{x}^2\left(x^2+x\right)-\frac{3}{4}x\left(x^3-1\right) \\ =\frac{3}{2}{x}^4-\frac{3}{2}{x}^2+\frac{1}{4}{x}^4+\frac{1}{4}{x}^3-\frac{3}{4}{x}^4+\frac{3}{4}x \\ =\frac{3}{2}{x}^4+\frac{1}{4}{x}^4-\frac{3}{4}{x}^4+\frac{1}{4}{x}^3-\frac{3}{2}{x}^2+\frac{3}{4}x \\ =\left(\frac{6+1-3}{4}\right)x^4+\frac{1}{4}{x}^3-\frac{3}{2}{x}^2+\frac{3}{4}x \\ =x^4+\frac{1}{4}{x}^3-\frac{3}{2}{x}^2+\frac{3}{4}x \end{gathered}$$

(xii) To simplify, we will use distributive law as follows:

$$\begin{gathered} a^{2}b\left(a-b^2\right)+a{b}^2\left(4ab-2a^2\right)-a^{3}b\left(1-2b\right) \\ =a^{3}b-a^2{b}^3+4a^2{b}^3-2a^3{b}^2-a^{3}b+2a^3{b}^2 \\ =a^{3}b-a^{3}b-a^2{b}^3+4a^2{b}^3-2a^3{b}^2+2a^3{b}^2 \\ =3a^2{b}^3 \end{gathered}$$

(xiii) To simplify, we will use distributive law as follows:​

$$\begin{gathered} a^{2}b\left(a^3-a+1\right)-ab\left(a^4-2a^2+2a\right)-b\left(a^3-a^2-1\right) \\ =a^{5}b-a^{3}b+a^{2}b-a^{5}b+2a^{3}b-2a^{2}b-a^{3}b+a^{2}b+b \\ =a^{5}b-a^{5}b-a^{3}b+2a^{3}b-a^{3}b+a^{2}b-2a^{2}b+a^{2}b+b \\ =b \end{gathered}$$