RD Sharma 2020 solution class 9 chapter 4 Algebraic Identities FBQS

FBQS

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Question 1:

If (a – b)² + (b – c)² + (c – a)² = 0, then __________.

Answer 1:

$$\begin{gathered} \mathrm{Given}: \\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0 \\ \Rightarrow {\left(a-b\right)}^2=0 \mathrm{and} {\left(b-c\right)}^2=0 \mathrm{and} {\left(c-a\right)}^2=0 \\ \Rightarrow \left(a-b\right)=0 \mathrm{and} \left(b-c\right)=0 \mathrm{and} \left(c-a\right)=0 \\ \Rightarrow a=b \mathrm{and} b=c \mathrm{and} c=a \\ \Rightarrow a=b=c \end{gathered}$$


Hence, if (a – b)² + (b – c)² + (c – a)² = 0, then a = b = c.

Question 2:

If $a+\frac{1}{a}=-2, \mathrm{then} a^2+\frac{1}{a^2}= \_\_\_\_\_\_\_\_\_.$

Answer 2:

$$\begin{gathered} \mathrm{Given}: \\ a+\frac{1}{a}=-2 \\ \mathrm{Now}, \\ a+\frac{1}{a}=-2 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(a+\frac{1}{a}\right)}^2={\left(-2\right)}^2 \\ \Rightarrow a^2+{\left(\frac{1}{a}\right)}^2+2a\frac{1}{a}=4 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow a^2+\frac{1}{a^2}+2=4 \\ \Rightarrow a^2+\frac{1}{a^2}=4-2 \\ \Rightarrow a^2+\frac{1}{a^2}=2 \end{gathered}$$

Hence, if $a+\frac{1}{a}=-2, \mathrm{then} a^2+\frac{1}{a^2}=\underline{2}.$

Question 3:

If $\frac{{37}^3-{28}^3}{{37}^2+37.28+{28}^2}=\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 3:

$$\begin{gathered} \frac{{37}^3-{28}^3}{{37}^2+37.28+{28}^2} \\ =\frac{\left(37-28\right)\left({37}^2+{28}^2+\left(37\right)\left(28\right)\right)}{{37}^2+37.28+{28}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)\right) \\ =\frac{\left(37-28\right)}{1} \\ =9 \end{gathered}$$

Hence, if $\frac{{37}^3-{28}^3}{{37}^2+37.28+{28}^2}=\underline{9}.$

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Question 4:

If a² – 2a – 1 = 0, then $a^2+\frac{1}{a^2}$ = __________.

Answer 4:

$$\begin{gathered} \mathrm{Given}: \\ {a}^2-2a-1=0 \\ \Rightarrow a^2-1=2a \\ \mathrm{Dividing} \text{'}a\text{'} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{a^2-1}{a}=\frac{2a}{a} \\ \Rightarrow \frac{a^2}{a}-\frac{1}{a}=2 \\ \Rightarrow a-\frac{1}{a}=2 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(a-\frac{1}{a}\right)}^2=2^2 \\ \Rightarrow a^2+{\left(\frac{1}{a}\right)}^2-2a\frac{1}{a}=4 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ \Rightarrow a^2+\frac{1}{a^2}-2=4 \\ \Rightarrow a^2+\frac{1}{a^2}=6 \end{gathered}$$

Hence, if a² – 2a – 1 = 0, then $a^2+\frac{1}{a^2}$ = 6.

Question 5:

If a² + b² + c² = 24 and ab + bc + ca = – 4, then a + b + c = ________.

Answer 5:

$$\begin{gathered} \mathrm{Given}: \\ {a}^2+b^2+c^2=24 ...\left(1\right) \\ ab+bc+ca=-4 ...\left(2\right) \\ \mathrm{Now}, \\ {\left(a+b+c\right)}^2 \\ =a^2+b^2+c^2+2\left(ab+bc+ca\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\right) \\ =24+2\left(-4\right) \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =24-8 \\ =16 \\ \mathrm{Since}, {\left(a+b+c\right)}^2=16 \\ \mathrm{Therefore}, a+b+c=\pm 4. \end{gathered}$$

Hence, if a² + b² + c² = 24 and ab + bc + ca = – 4, then a + b + c = ±4.

Question 6:

If x + y + z = 0, then (x + y)³ + (y + z)³ + (z + x)³ = ________.

Answer 6:

$$\begin{gathered} \mathrm{Given}: \\ x+y+z=0 \\ \Rightarrow x+y=-z ...\left(1\right) \\ \Rightarrow z+y=-x ...\left(2\right) \\ \Rightarrow x+z=-y ...\left(3\right) \\ \mathrm{Now}, \\ {\left(x+y\right)}^3+{\left(y+z\right)}^3+{\left(z+x\right)}^3 \\ ={\left(-z\right)}^3+{\left(-x\right)}^3+{\left(-y\right)}^3 \left(\mathrm{From} \left(1\right), \left(2\right) \mathrm{and} \left(3\right)\right) \\ =-z^3-x^3-y^3 \\ =-\left(z^3+x^3+y^3\right) \\ =-3xyz \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: \mathrm{if} \left(a+b+c\right)=0 \mathrm{then} a^3+b^3+c^3=3abc\right) \end{gathered}$$

Hence, if x + y + z = 0, then (x + y)³ + (y + z)³ + (z + x)³ = −3xyz.

Question 7:

If x + y + z = 5 and xy + yz + zx = 7, then x³ + y³ + z³ – 3xyz = _________.

Answer 7:

$$\begin{gathered} \mathrm{Given}: \\ x+y+z=5 ...\left(1\right) \\ xy+yz+zx=7 ...\left(2\right) \\ {\left(x+y+z\right)}^2=x^2+y^2+z^2+2xy+2yz+2zx \\ \Rightarrow 5^2=x^2+y^2+z^2+2\left(7\right) \\ \Rightarrow 25=x^2+y^2+z^2+14 \\ \Rightarrow x^2+y^2+z^2=11 ...\left(3\right) \\ \mathrm{Now}, \\ {x}^3+y^3+z^3-3xyz \\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right) \\ =\left(5\right)\left(11-7\right) \left(\mathrm{From} \left(1\right), \left(2\right) \mathrm{and} \left(3\right)\right) \\ =5\times 4 \\ =20 \end{gathered}$$

Hence, if x + y + z = 5 and xy + yz + zx = 7, then x³ + y³ + z³ – 3xyz = 20.

Question 8:

If (a + b + c) {(a – b)² + (b – c)² + (c – a)²} = k(a³ + b³ + c³ – 3abc), then k = ________.

Answer 8:

$$\begin{gathered} \mathrm{Given}: \\ \left(a+b+c\right)\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\}=k\left(a^3+b^3+c^3-3abc\right) \\ \left(a+b+c\right)\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\}=k\left(a^3+b^3+c^3-3abc\right) \\ \Rightarrow \left(a+b+c\right)\left\{a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac\right\}=k\left(a^3+b^3+c^3-3abc\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(x-y\right)}^2=x^2+y^2-2xy\right) \\ \Rightarrow \left(a+b+c\right)\left\{2a^2+2b^2+2c^2-2ab-2bc-2ac\right\}=k\left(a^3+b^3+c^3-3abc\right) \\ \Rightarrow \left(a+b+c\right)\left\{2a^2+2b^2+2c^2-2ab-2bc-2ac\right\}=k\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right) \\ \Rightarrow 2\left(a+b+c\right)\left\{a^2+b^2+c^2-ab-bc-ac\right\}=k\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right) \\ \Rightarrow k=2 \end{gathered}$$

Hence, if (a + b + c) {(a – b)² + (b – c)² + (c – a)²} = k(a³ + b³ + c³ – 3abc), then k = 2.

Question 9:

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \mathrm{and} abc=2, \mathrm{then} a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c=\_\_\_\_\_\_\_\_\_.$

Answer 9:

$$\begin{gathered} \mathrm{Given}: \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 ...\left(1\right) \\ abc=2 ...\left(2\right) \\ \mathrm{Now}, \\ a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c \\ =a^2{b}^2{c}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ ={\left(abc\right)}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ ={\left(2\right)}^2\left(1\right) \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =4 \end{gathered}$$

Hence, if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \mathrm{and} abc=2, \mathrm{then} a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c=\underline{4}.$

Question 10:

If $a+b+c=6, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}, \mathrm{then} \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\_\_\_\_\_\_\_\_\_\_.$

Answer 10:

$$\begin{gathered} \mathrm{Given}: \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2} ...\left(1\right) \\ a+b+c=6 ...\left(2\right) \\ \mathrm{Now}, \\ \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b} \\ \mathrm{Adding} \mathrm{and} \mathrm{subtracting} 3, \mathrm{we} \mathrm{get} \\ =\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3-3 \\ =\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1+1+1-3 \\ =\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{a}{a}+\frac{b}{b}+\frac{c}{c}-3 \\ =\left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-3 \\ =\left(\frac{a+b+c}{a}\right)+\left(\frac{a+b+c}{b}\right)+\left(\frac{a+b+c}{c}\right)-3 \\ =\left(\frac{6}{a}\right)+\left(\frac{6}{b}\right)+\left(\frac{6}{c}\right)-3 \left(\mathrm{From} \left(2\right)\right) \\ =\frac{6}{a}+\frac{6}{b}+\frac{6}{c}-3 \\ =6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ =6\left(\frac{3}{2}\right)-3 \left(\mathrm{From} \left(1\right)\right) \\ =9-3 \\ =6 \end{gathered}$$

Hence, if $a+b+c=6, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}, \mathrm{then} \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\underline{6}.$

Question 11:

If $\frac{a}{b}+\frac{b}{a}=2, \mathrm{then} {\left(\frac{a}{b}\right)}^{100}-{\left(\frac{b}{a}\right)}^{100}=\_\_\_\_\_\_\_\_\_\_.$

Answer 11:

$$\begin{gathered} \mathrm{Given}: \\ \frac{a}{b}+\frac{b}{a}=2 \\ \Rightarrow \frac{a^2+b^2}{ab}=2 \\ \Rightarrow a^2+b^2=2ab \\ \Rightarrow a^2+b^2-2ab=0 \\ \Rightarrow {\left(a-b\right)}^2=0 \\ \Rightarrow \left(a-b\right)=0 \\ \Rightarrow a=b ...\left(1\right) \\ \mathrm{Now}, \\ {\left(\frac{a}{b}\right)}^{100}-{\left(\frac{b}{a}\right)}^{100}={\left(\frac{a}{a}\right)}^{100}-{\left(\frac{a}{a}\right)}^{100} \\ =1^{100}-1^{100} \\ =0 \end{gathered}$$

Hence, if $\frac{a}{b}+\frac{b}{a}=2, \mathrm{then} {\left(\frac{a}{b}\right)}^{100}-{\left(\frac{b}{a}\right)}^{100}=\underline{0}.$

Question 12:

If $x^2+y^2-xy=3 \mathrm{and} y-x=1, \mathrm{then}\frac{xy}{x^2+y^2}=\_\_\_\_\_\_\_\_\_\_\_.$

Answer 12:

$$\begin{gathered} \mathrm{Given}: \\ {x}^2+y^2-xy=3 ...\left(1\right) \\ y-x=1 ...\left(2\right) \\ \mathrm{Now}, \\ y-x=1 \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow {\left(y-x\right)}^2=1^2 \\ \Rightarrow x^2+y^2-2xy=1 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ \Rightarrow x^2+y^2-xy-xy=1 \\ \Rightarrow 3-xy=1 \left(\mathrm{From} \left(1\right)\right) \\ \Rightarrow xy=2 ...\left(3\right) \\ \mathrm{Thus}, \\ \frac{xy}{x^2+y^2}=\frac{2}{3+xy} \left(\mathrm{From} \left(1\right) \mathrm{and} \left(3\right)\right) \\ =\frac{2}{3+2} \\ =\frac{2}{5} \end{gathered}$$

Hence, if $x^2+y^2-xy=3 \mathrm{and} y-x=1, \mathrm{then}\frac{xy}{x^2+y^2}=\underline{\frac{2}{5}}.$