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Question: Using suitable identities, expand the following: (a) \({{\left( \dfrac{x}{5}-3y \right)}^{2}}\) ...

Using suitable identities, expand the following:
(a) (x5−3y)2{{\left( \dfrac{x}{5}-3y \right)}^{2}}
(b) (11x+0.2y)2{{\left( 11x+0.2y \right)}^{2}}
(c) (4a−56)2{{\left( 4a-56 \right)}^{2}}
(d) (y−25x)3{{\left( y-\dfrac{2}{5}x \right)}^{3}}
(e) (−3a+5b+4c)2{{\left( -3a+5b+4c \right)}^{2}}
(f) (12a−b−13c)\left( \dfrac{1}{2}a-b-\dfrac{1}{3}c \right)

Explanation

Solution

Hint: To expand (a) and (c), use the identity ′′(a−b)2=a2−2ab+b2′′''{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}''. To expand (b), use the identity ′′(a+b)2=a2+2ab+b2′′''{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}''. To expand (d), use the identity ′′(a−b)3=a3−b3−3ab(a−b)′′''{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)''. To expand (e) and (f), use the identity ′′(a+b+c)3=a2+b2+c2+2ab+2bc+2ca′′''{{\left( a+b+c \right)}^{3}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca''.

Complete step-by-step answer:
(a) (x5−3y)2{{\left( \dfrac{x}{5}-3y \right)}^{2}}
We have to expand this expression using a suitable identity.
Identity that can be used: ′′(a−b)2=a2−2ab+b2′′''{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}''
Here a=x5a=\dfrac{x}{5} and b=3yb=3y .
Putting a=x5 and b=3ya=\dfrac{x}{5}\,and\,b=3y in the above identity, we will get.
(x5−3y)2=(x5)2−2(x5)(3y)+(3y)2 ⇒(x5−3y)2=x225−6xy5+9y2 \begin{aligned} & {{\left( \dfrac{x}{5}-3y \right)}^{2}}={{\left( \dfrac{x}{5} \right)}^{2}}-2\left( \dfrac{x}{5} \right)\left( 3y \right)+{{\left( 3y \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{x}{5}-3y \right)}^{2}}=\dfrac{{{x}^{2}}}{25}-\dfrac{6xy}{5}+9{{y}^{2}} \\\ \end{aligned}
(b) (11x+0.2y)2{{\left( 11x+0.2y \right)}^{2}}
To expand this algebraic expression, the identity that can be used is: ′′(a+b)2=a2+2ab+b2′′''{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}''.
Here a=11x and b=0.2y
On putting a=11x and b=0.2y in the above identity, we will get: -
(11x+0.2y)2=(11x)2+2(11x)(0.2y)+(0.2y)2{{\left( 11x+0.2y \right)}^{2}}={{\left( 11x \right)}^{2}}+2\left( 11x \right)\left( 0.2y \right)+{{\left( 0.2y \right)}^{2}}
⇒(11x+0.2y)2=121x2+4.4xy+0.04y2\Rightarrow {{\left( 11x+0.2y \right)}^{2}}=121{{x}^{2}}+4.4xy+0.04{{y}^{2}}
(c)(4a−56)2{{\left( 4a-56 \right)}^{2}}
To expand the algebraic expression, the identity that can be used is: ′′(x−y)2=x2−2xy+y2′′''{{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}'' on putting x=4a and y=5b, we will get-
(4a−5b)2=(4a)2−2(4a)(5b)+(5b)2 ⇒(4a−5b)2=16a2−40ab+25b2 \begin{aligned} & {{\left( 4a-5b \right)}^{2}}={{\left( 4a \right)}^{2}}-2\left( 4a \right)\left( 5b \right)+{{\left( 5b \right)}^{2}} \\\ & \Rightarrow {{\left( 4a-5b \right)}^{2}}=16{{a}^{2}}-40ab+25{{b}^{2}} \\\ \end{aligned}
(d) (y−25x)3{{\left( y-\dfrac{2}{5}x \right)}^{3}}
To expand this algebraic expression, the identity that can be used is: ′′(a−b)3=a3−b3−3ab(a−b)′′''{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)''. On putting a=y and b= 2x5\dfrac{2x}{5} in the above identity, we will ge (y−25x)3=y3−(2x5)3−3y(2x5)(y−2x5) ⇒(y−25x)3=y3−(8x125)3−(6xy5)(y−2x5) \begin{aligned} & \\\ & {{\left( y-\dfrac{2}{5}x \right)}^{3}}={{y}^{3}}-{{\left( \dfrac{2x}{5} \right)}^{3}}-3y\left( \dfrac{2x}{5} \right)\left( y-\dfrac{2x}{5} \right) \\\ & \Rightarrow {{\left( y-\dfrac{2}{5}x \right)}^{3}}={{y}^{3}}-{{\left( \dfrac{8x}{125} \right)}^{3}}-\left( \dfrac{6xy}{5} \right)\left( y-\dfrac{2x}{5} \right) \\\ \end{aligned}
(e)(−3a+5b+4c)2{{\left( -3a+5b+4c \right)}^{2}}
To expand this algebraic expression, the identity that can be used is: ′′(x+y+z)3=x2+y2+z2+2xy+2yz+2zx′′''{{\left( x+y+z \right)}^{3}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx'' . on putting x=-34, y=5b and z=4c in above identity, we will get-

& {{\left( -3a+5b+4c \right)}^{3}}={{\left( -3a \right)}^{2}}+{{\left( 5b \right)}^{2}}+{{\left( 4c \right)}^{2}}+2\left( -3a \right)\left( 5b \right)+2\left( 5b \right)\left( 4c \right)+2\left( -3a \right)\left( 4c \right) \\\ & \Rightarrow {{\left( -3a+5b+4c \right)}^{3}}=9{{a}^{2}}+25{{b}^{2}}+16{{c}^{2}}-30ab+40bc-24ac \\\ \end{aligned}$$$$$$ (f) $\left( \dfrac{1}{2}a-b-\dfrac{1}{3}c \right)$ To expand the above algebraic expression, the identity that can be used is $''{{\left( x+y+z \right)}^{3}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx''$. On putting x= $\dfrac{a}{2}$, y=-b and z= $\dfrac{-c}{3}$, we get- $$\begin{aligned} & {{\left( \dfrac{1}{2}a-b-\dfrac{1}{3}c \right)}^{3}}={{\left( \dfrac{1}{2}a \right)}^{2}}+{{\left( -b \right)}^{2}}+{{\left( -\dfrac{1}{3}c \right)}^{2}}+2\left( \dfrac{1}{2}a \right)\left( -b \right)+2\left( -b \right)\left( -\dfrac{c}{3} \right)+2\left( \dfrac{a}{2} \right)\left( -\dfrac{c}{3} \right) \\\ & \Rightarrow \left( \dfrac{1}{2}a-b-\dfrac{1}{3}c \right)=\dfrac{{{a}^{2}}}{4}+{{b}^{2}}+\dfrac{{{c}^{2}}}{9}-ab+\dfrac{2bc}{3}-\dfrac{ac}{3} \\\ \end{aligned}$$ Note: While using the identities do not get confused in the sign. Students generally get confused and make sign mistakes which leads to wrong answers. Such confusion should be avoided and the identities must be memorized correctly.