RS Aggarwal 2019,2020 solution class 7 chapter 6 Algebraic Expressions Exercise 6C

Exercise 6C

Page-104

Question 1:

Find the product:

4a(3a + 7b)

Answer 1:

$$\begin{gathered} =4\mathrm{a}\times 3\mathrm{a} + 4\mathrm{a}\times 7\mathrm{b} \\ =4\times 3 \times {\mathrm{a}}^{(1+1)} + 4\times 7 \times \mathrm{a}\times \mathrm{b} \\ =12{\mathrm{a}}^2 +28\mathrm{ab} \end{gathered}$$

Question 2:

Find the product:

5a(6a − 3b)

Answer 2:

$$\begin{gathered} =5\mathrm{a}\times 6\mathrm{a} -5\mathrm{a}\times 3\mathrm{b} \\ =5\times 6\times \mathrm{a}\times \mathrm{a} -(5\times 3\times \mathrm{a}\times \mathrm{b}) \\ =30{\mathrm{a}}^2-15\mathrm{ab} \end{gathered}$$

Question 3:

Find the product:

8a²(2a + 5b)

Answer 3:

$$\begin{gathered} =8{\mathrm{a}}^2\times 2\mathrm{a} +8{\mathrm{a}}^2\times 5\mathrm{b} \\ =8\times 2\times {\mathrm{a}}^2\times \mathrm{a} +8\times 5\times {\mathrm{a}}^2\times \mathrm{b} \\ =16{\mathrm{a}}^{(2+1)} +40{\mathrm{a}}^2\mathrm{b} \\ =16{\mathrm{a}}^3+40{\mathrm{a}}^2\mathrm{b} \end{gathered}$$

Question 4:

Find the product:

9x²(5x + 7)

Answer 4:

$$\begin{gathered} =9{\mathrm{x}}^2\times 5\mathrm{x} +9{\mathrm{x}}^2\times 7 \\ =9\times 5\times {\mathrm{x}}^2\times \mathrm{x} + 9\times 7\times {\mathrm{x}}^2 \\ =45{\mathrm{x}}^{(2+1)} +63{\mathrm{x}}^2 \\ =45{\mathrm{x}}^3+63{\mathrm{x}}^2 \end{gathered}$$

Question 5:

Find the product:

ab(a² − b²)

Answer 5:

$$\begin{gathered} =\mathrm{ab}\times {\mathrm{a}}^2-\mathrm{ab}\times {\mathrm{b}}^2 \\ ={\mathrm{a}}^{(1+2)}\mathrm{b}-{\mathrm{ab}}^{(1+2)} \\ ={\mathrm{a}}^3\mathrm{b} -{\mathrm{ab}}^3 \end{gathered}$$

Question 6:

Find the product:

2x²(3x − 4x²)

Answer 6:

$$\begin{gathered} =2{\mathrm{x}}^2\times 3\mathrm{x} -2{\mathrm{x}}^2\times 4{\mathrm{x}}^2 \\ =2\times 3\times {\mathrm{x}}^2\times \mathrm{x} -2\times 4\times {\mathrm{x}}^2\times {\mathrm{x}}^2 \\ =6\times {\mathrm{x}}^{(2+1)} -8\times {\mathrm{x}}^{(2+2)} \\ =6{\mathrm{x}}^3-8{\mathrm{x}}^4 \end{gathered}$$

Question 7:

Find the product:

$\frac{3}{5}{m}^{2}n(m+5n)$

Answer 7:

$$\begin{gathered} =\frac{3}{5}{\mathrm{m}}^2\mathrm{n} \times \mathrm{m} +\frac{3}{5}{\mathrm{m}}^2\mathrm{n}\times 5\mathrm{n} \\ =\frac{3}{5}\times {\mathrm{m}}^2\times \mathrm{m}\times \mathrm{n}+\frac{3}{5}\times 5\times {\mathrm{m}}^2\times \mathrm{n}\times \mathrm{n} \\ =\frac{3}{5}{\mathrm{m}}^{(2+1)}\times \mathrm{n} +3\times {\mathrm{m}}^2\times {\mathrm{n}}^{(1+1)} \\ =\frac{3}{5}{\mathrm{m}}^3\mathrm{n} +3{\mathrm{m}}^2{\mathrm{n}}^2 \end{gathered}$$

Question 8:

Find the product:

−17x²(3x − 4)

Answer 8:

$$\begin{gathered} =-17{\mathrm{x}}^2\times 3\mathrm{x} -(-17{\mathrm{x}}^2\times 4) \\ =-17\times 3\times {\mathrm{x}}^2\times \mathrm{x} +17\times 4\times {\mathrm{x}}^2 \\ =-51\times {\mathrm{x}}^{(2+1)} +68{\mathrm{x}}^2 \\ =-51{\mathrm{x}}^3 +68{\mathrm{x}}^2 \end{gathered}$$

Question 9:

Find the product:

$\frac{7}{2}{x}^2\left(\frac{4}{7}x+2\right)$

Answer 9:

$$\begin{gathered} =\frac{7}{2}{\mathrm{x}}^2\times \frac{4}{7}\times \mathrm{x} +\frac{7}{2}{\mathrm{x}}^2\times 2 \\ =\frac{7}{2}\times \frac{4}{7}\times {\mathrm{x}}^2\times \mathrm{x} +\frac{7}{2}\times 2\times {\mathrm{x}}^2 \\ =2\times {\mathrm{x}}^{(2+1)}+7{\mathrm{x}}^2 \\ =2{\mathrm{x}}^3 +7{\mathrm{x}}^2 \end{gathered}$$

Question 10:

Find the product:

−4x²y(3x² − 5y)

Answer 10:

$$\begin{gathered} =-4{\mathrm{x}}^2\mathrm{y} \times 3{\mathrm{x}}^2 -(-4{\mathrm{x}}^2\mathrm{y}\times 5\mathrm{y}) \\ =-4\times 3\times {\mathrm{x}}^2\times {\mathrm{x}}^2\times \mathrm{y} + 4\times 5\times {\mathrm{x}}^2\times \mathrm{y}\times \mathrm{y} \\ =-12\times {\mathrm{x}}^{(2+2)}\times \mathrm{y} +20\times {\mathrm{x}}^2\times {\mathrm{y}}^{(1+1)} \\ =-12{\mathrm{x}}^4\mathrm{y} +20{\mathrm{x}}^2{\mathrm{y}}^2 \end{gathered}$$

Question 11:

Find the product:

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

Answer 11:

$$\begin{gathered} =\frac{-4}{27}\mathrm{xyz}\times \frac{9}{2}{\mathrm{x}}^2\mathrm{yz} -(\frac{-4}{27}\mathrm{xyz} \times \frac{3}{4}{\mathrm{xyz}}^2) \\ =\frac{-4}{27}\times \frac{9}{2}\times \mathrm{x}\times {\mathrm{x}}^2\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z} + \frac{4}{27}\times \frac{3}{4}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times {\mathrm{z}}^2 \\ =\frac{-2}{3}\times {\mathrm{x}}^{(1+2) }\times {\mathrm{y}}^{(1+1)}\times {\mathrm{z}}^{(1+1)} + \frac{1}{9}\times {\mathrm{x}}^{(1+1)}\times {\mathrm{y}}^{(1+1)}\times {\mathrm{z}}^{(1+2)} \\ =\frac{-2}{3}{\mathrm{x}}^3{\mathrm{y}}^2{\mathrm{z}}^2 +\frac{1}{9}{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^3 \end{gathered}$$

Question 12:

Find the product:

9t²(t + 7t³)

Answer 12:

$$\begin{gathered} =9{\mathrm{t}}^2\times \mathrm{t} +9{\mathrm{t}}^2\times 7{\mathrm{t}}^3 \\ =9\times {\mathrm{t}}^2\times \mathrm{t}+9\times 7\times {\mathrm{t}}^2\times {\mathrm{t}}^3 \\ =9\times {\mathrm{t}}^{(2+1)} +63\times {\mathrm{t}}^{(2+3)} \\ =9{\mathrm{t}}^3+63{\mathrm{t}}^5 \end{gathered}$$

Question 13:

Find the product:

10a²(0.1a − 0.5b)

Answer 13:

$$\begin{gathered} =10a^2\times 0.1a - 10a^2\times 0.5b \\ =10\times 0.1\times a^2\times a -10\times 0.5\times a^2\times b \\ =1\times a^{(2+1)} -5 a^{2}b \\ =a^3 -5a^{2}b \end{gathered}$$

Question 14:

Find the product:

1.5a(10a² − 100ab²)

Answer 14:

$$\begin{gathered} =1.5\mathrm{a}\times 10{\mathrm{a}}^2\mathrm{b} -1.5\mathrm{a}\times 100{\mathrm{ab}}^2 \\ =1.5\times 10\times \mathrm{a}\times {\mathrm{a}}^2\mathrm{b} - 1.5\times 100\times \mathrm{a}\times \mathrm{a}\times {\mathrm{b}}^2 \\ =15\times {\mathrm{a}}^{(1+2)}\mathrm{b}-150 \times {\mathrm{a}}^{(1+1)}\times {\mathrm{b}}^2 \\ =15{\mathrm{a}}^3\mathrm{b}-150{\mathrm{a}}^2{\mathrm{b}}^2 \end{gathered}$$

Question 15:

Find the product:

$\frac{2}{3}abc(a^2+b^2-3c^2)$

Answer 15:

$$\begin{gathered} =\frac{2}{3}\mathrm{abc}\times {\mathrm{a}}^2 +\frac{2}{3}\mathrm{abc}\times {\mathrm{b}}^2-\frac{2}{3}\mathrm{abc}\times 3{\mathrm{c}}^2 \\ =\frac{2}{3}\mathrm{a}\times {\mathrm{a}}^2\times \mathrm{b}\times \mathrm{c}+\frac{2}{3}\mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{c} -\frac{2}{3}\times 3\times \mathrm{a}\times \mathrm{b}\times \mathrm{c}\times {\mathrm{c}}^2 \\ =\frac{2}{3}\times {\mathrm{a}}^{(1+2)}\times \mathrm{b}\times \mathrm{c}+\frac{2}{3}\times \mathrm{a}\times {\mathrm{b}}^{(1+2)}\times \mathrm{c} -2\times \mathrm{a}\times \mathrm{b}\times {\mathrm{c}}^{(1+2)} \\ =\frac{2}{3}{\mathrm{a}}^3\mathrm{bc} +\frac{2}{3}{\mathrm{ab}}^3\mathrm{c} -2{\mathrm{abc}}^3 \end{gathered}$$

Question 16:

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

Answer 16:

$$\begin{gathered} {\mathrm{24x}}^2(1-2x) \\ =24{\mathrm{x}}^2\times 1 -24{\mathrm{x}}^2\times 2\mathrm{x} \\ =24{\mathrm{x}}^2 -24\times 2\times {\mathrm{x}}^2\times \mathrm{x} \\ =24{\mathrm{x}}^2-48{\mathrm{x}}^3 \\ \mathrm{W}\mathrm{h}\mathrm{e}\mathrm{n} \mathrm{x} =2: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. = 24{\mathrm{x}}^2(1-2\mathrm{x}) = 24\times 2^2(1-2\times 2) = 96 (1-4)=96\times (-3) = -288 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= 24{\mathrm{x}}^2-48{\mathrm{x}}^2 = 24\times 2^2 - 48\times 2^3 = 96-384 = -288 \\ \hspace{0.17em}\mathrm{L}.\mathrm{H}.\mathrm{S}.= \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore 24{\mathrm{x}}^2(1-2\mathrm{x}) = 24{\mathrm{x}}^2-48{\mathrm{x}}^3 \end{gathered}$$

Question 17:

Find the product ab(a²+b²) and evaluate it for a = 2 and b = $\frac{1}{2}$.

Answer 17:

$$\begin{gathered} \mathrm{ab}{(}^2{+b}^2) \\ =\mathrm{a}\mathrm{b}\times {\mathrm{a}}^2+\mathrm{a}\mathrm{b}\times {\mathrm{b}}^2 \\ =\mathrm{a}\times {\mathrm{a}}^2\times \mathrm{b} +\mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2 \\ ={\mathrm{a}}^{(1+2)}\times \mathrm{b} +\mathrm{a}\times {\mathrm{b}}^{(1+2)} \\ ={\mathrm{a}}^3\mathrm{b} +\mathrm{a}{\mathrm{b}}^3 \\ \mathrm{W}\mathrm{h}\mathrm{e}\mathrm{n} \mathrm{a}=2 \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{b} =\frac{1}{2}, \mathrm{w}\mathrm{e} \mathrm{g}\mathrm{e}\mathrm{t}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. = \mathrm{a}\mathrm{b}({\mathrm{a}}^2 +{\mathrm{b}}^2) = 2\times \frac{1}{2}(2^2+\frac{1}{2^2}) = 4 +\frac{1}{4}=\frac{17}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}. = {\mathrm{a}}^3\mathrm{b} +\mathrm{a}{\mathrm{b}}^3 = 2^3\times \frac{1}{2} + 2\times {\left(\frac{1}{2}\right)}^3 = 4+ \frac{1}{4} =\frac{17}{4} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}. = \mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 18:

Find the product s (s² − st) and find its value for s = 2 and t = 3.

Answer 18:

$$\begin{gathered} {\mathrm{s\; (s}}^2-\; st) \\ =\mathrm{s}\times {\mathrm{s}}^2-\mathrm{s}\times \mathrm{st} \\ ={\mathrm{s}}^{(1+2)} -{\mathrm{s}}^{(1+1)}\times \mathrm{t} \\ ={\mathrm{s}}^3-{\mathrm{s}}^2\mathrm{t} \\ \mathrm{When} \mathrm{s} =2 \mathrm{and} \mathrm{t} = 3, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \mathrm{s}({\mathrm{s}}^{2 }-\mathrm{st}) = 2(2^2-2\times 3) = 2 \times (4-6) = -4 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}. = {\mathrm{s}}^3-{\mathrm{s}}^2\mathrm{t} = 2^3-2^2\times 3 = 8 - 12 = -4 \\ \hspace{0.17em}\mathrm{L}.\mathrm{H}.\mathrm{S}.= \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore \mathrm{s}({\mathrm{s}}^{2 }-\mathrm{st}) = {\mathrm{s}}^3-{\mathrm{s}}^2\mathrm{t} \end{gathered}$$

Question 19:

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

Answer 19:

$$\begin{gathered} -3\mathrm{y}(\mathrm{xy}+{\mathrm{y}}^2) \\ =-3\mathrm{y}\times \mathrm{xy} -3\mathrm{y}\times {\mathrm{y}}^2 \\ =-3\times \mathrm{x}\times \mathrm{y}\times \mathrm{y} -3\times \mathrm{y}\times {\mathrm{y}}^2 \\ =-3\times \mathrm{x}\times {\mathrm{y}}^{(1+1)} -3\times {\mathrm{y}}^{(1+2)} \\ =-3{\mathrm{xy}}^{2 } -3{\mathrm{y}}^3 \\ \mathrm{When} \mathrm{x} = 4 \mathrm{and} \mathrm{y} =5, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= -3\mathrm{y}(\mathrm{xy} +{\mathrm{y}}^2) = -3\times 5(4\times 5 + 5^2) = -15 \times (20 +25) = -675 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}. =-3{\mathrm{xy}}^{2 } -3{\mathrm{y}}^3 = -3\times 4\times 5^2 -3\times 5^3 = -300- 375 =-675 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore -3\mathrm{y}(\mathrm{xy} +{\mathrm{y}}^2) = -3{\mathrm{xy}}^{2 } -3{\mathrm{y}}^3 \end{gathered}$$

Question 20:

Simplify

a(b − c) + b(c − a) + c(a − b)

Answer 20:

$$\begin{gathered} \mathrm{a(b\; -\; c)\; +\; b(c\; -\; a)\; +\; c(a\; -\; b)} \\ =\mathrm{a}\times \mathrm{b} -\mathrm{a}\times \mathrm{c} +\mathrm{b}\times \mathrm{c} -\mathrm{b}\times \mathrm{a} +\mathrm{c}\times \mathrm{a} -\mathrm{c}\times \mathrm{b} \\ =\mathrm{a}\mathrm{b} -\mathrm{a}\mathrm{c} +\mathrm{b}\mathrm{c} -\mathrm{a}\mathrm{b} +\mathrm{a}\mathrm{c} -\mathrm{b}\mathrm{c} \\ =0 \end{gathered}$$

Question 21:

Simplify

a(b − c) − b(c − a) − c(a − b)

Answer 21:

$$\begin{gathered} \mathrm{a}(\mathrm{b}-\mathrm{c})-\mathrm{b} (\mathrm{c}-\mathrm{a})-\mathrm{c}(\mathrm{a}-\mathrm{b}) \\ =\mathrm{a}\times \mathrm{b} -\mathrm{a}\times \mathrm{c} -\mathrm{b}\times \mathrm{c}+\mathrm{b}\times \mathrm{a} -\mathrm{c}\times \mathrm{a} +\mathrm{c}\times \mathrm{b} \\ =\mathrm{ab} +\mathrm{ab} -\mathrm{ac} - \mathrm{ac} -\mathrm{bc} +\mathrm{bc} \\ =2\mathrm{ab} - 2\mathrm{ac} \\ =2\mathrm{a}(\mathrm{b}-\mathrm{c}) \end{gathered}$$

Question 22:

Simplify

3x² + 2(x + 2) −3x(2x + 1)

Answer 22:

$$\begin{gathered} 3{\mathrm{x}}^2 +2(\mathrm{x}+2)-3\mathrm{x}(2\mathrm{x}+1) \\ =3{\mathrm{x}}^2 +2\times \mathrm{x} +2\times 2 -3\mathrm{x}\times 2\mathrm{x} -3\mathrm{x} \\ =3{\mathrm{x}}^2 +2\mathrm{x} +4 -6{\mathrm{x}}^2 -3\mathrm{x} \\ =-3{\mathrm{x}}^2-\mathrm{x} +4 \end{gathered}$$

Question 23:

Simplify

x(x + 4) + 3x(2x² − 1) + 4x² + 4

Answer 23:

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

Question 24:

Simplify

2x² + 3x(1 − 2x³) + x(x + 1)

Answer 24:

$$\begin{gathered} 2{\mathrm{x}}^2 +3\mathrm{x}(1-2{\mathrm{x}}^3)+\mathrm{x}(\mathrm{x}+1) \\ =2{\mathrm{x}}^2 +3\mathrm{x} -3\mathrm{x}\times 2{\mathrm{x}}^3 +{\mathrm{x}}^2 +\mathrm{x} \\ =2{\mathrm{x}}^2 +3\mathrm{x} -6\times {\mathrm{x}}^{(1+3)} +{\mathrm{x}}^2+\mathrm{x} \\ =2{\mathrm{x}}^2+3\mathrm{x} -6{\mathrm{x}}^4 +{\mathrm{x}}^2 +\mathrm{x} \\ =-6{\mathrm{x}}^4+3{\mathrm{x}}^2 +4\mathrm{x} \end{gathered}$$

Question 25:

Simplify

a²b(a − b²) + ab²(4ab − 2a²) −a³b(1 − 2b)

Answer 25:

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

Question 26:

Simplify

4st(s − t) −6s²(t − t²) −3t² (2s² − s) +2st (s − t)

Answer 26:

$$\begin{gathered} 4\mathrm{st}(\mathrm{s}-\mathrm{t})-6{\mathrm{s}}^2(\mathrm{t}-{\mathrm{t}}^2)-3{\mathrm{t}}^2(2{\mathrm{s}}^2-\mathrm{s})+2\mathrm{st}(\mathrm{s}-\mathrm{t}) \\ =4\mathrm{st}\times \mathrm{s} -4\mathrm{st}\times \mathrm{t} -6{\mathrm{s}}^2\times \mathrm{t}-6{\mathrm{s}}^2\times (-{\mathrm{t}}^2) -3{\mathrm{t}}^2\times 2{\mathrm{s}}^2-3{\mathrm{t}}^2\times (-\mathrm{s}) +2\mathrm{st}\times \mathrm{s} -2\mathrm{st}\times \mathrm{t} \\ =4\times {\mathrm{s}}^{(1+1)}\times \mathrm{t} -4\times \mathrm{s}\times {\mathrm{t}}^{(1+1)} -6{\mathrm{s}}^2\mathrm{t} +6{\mathrm{s}}^2{\mathrm{t}}^{2 }-6{\mathrm{t}}^2{\mathrm{s}}^2+3{\mathrm{t}}^2\mathrm{s} +2\times {\mathrm{s}}^{(1+1)}\times \mathrm{t} -2\times \mathrm{s}\times {\mathrm{t}}^{(1+1)} \\ =4{\mathrm{s}}^2\mathrm{t} -4{\mathrm{st}}^2-6{\mathrm{s}}^2\mathrm{t} +6{\mathrm{s}}^2{\mathrm{t}}^2-6{\mathrm{t}}^2{\mathrm{s}}^2+3{\mathrm{t}}^2\mathrm{s}+2{\mathrm{s}}^2\mathrm{t} -2{\mathrm{st}}^2 \\ =4{\mathrm{s}}^2\mathrm{t} -6{\mathrm{s}}^2\mathrm{t}+2{\mathrm{s}}^2\mathrm{t} -4{\mathrm{st}}^2 +3{\mathrm{st}}^2-2{\mathrm{st}}^2 \\ =-3{\mathrm{st}}^2 \end{gathered}$$