#### Explain Solution R D Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 28 maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$

Hint First we will try to make some elements of $R_{1} \text { or } C_{1} \quad(a+b)^{2}=a^{2}+2 a b+b^{2}$

Because from R.H.S \begin{aligned} &\left(a^{3}+b^{3}\right)^{2}=\left[(a+b)\left(a^{2}-a b+b^{2}\right)\right]^{2} \\ &=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)^{2} \end{aligned}

Given: $\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} a^{2}+b^{2}+2 a b & 2 a b+a^{2}+b^{2} & b^{2}+2 a b+a^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right| \\ &\because(a+b)^{2}=a^{2}+2 a b+b^{2} \\ &=\left|\begin{array}{ccc} (a+b)^{2} & (a+b)^{2} & (a+b)^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking }(a+b)^{2} \text { common from } \mathrm{R}_{1}\\ &=(a+b)^{2}\left|\begin{array}{ccc} 1 & 1 & 1 \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b)^{2}\left[0\left|\begin{array}{cc} a^{2}-2 a b & 2 a b \\ b^{2}-a^{2} & a^{2} \end{array}\right|-0\left|\begin{array}{cc} b^{2}-a^{2} & 2 a b \\ 2 a b-b^{2} & a^{2} \end{array}\right|+1\left|\begin{array}{cc} b^{2}-a^{2} & a^{2}-2 a b \\ 2 a b-b^{2} & b^{2}-a^{2} \end{array}\right|\right]\\ &=(a+b)^{2}\left[0-0+1\left\{\left(b^{2}-a^{2}\right)\left(b^{2}-a^{2}\right)\right\}-\left(a^{2}-2 a b\right)\left(2 a b-b^{2}\right)\right]\\ &=(a+b)^{2}\left(b^{4}+a^{4}-2 a^{2} b^{2}-2 a^{3} b+a^{2} b^{2}+4 a^{2} b^{2}-2 a b^{3}\right) \end{aligned}

\begin{aligned} &=(a+b)^{2}\left[\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+2 a^{2} b^{2}-2 a^{3} b-2 a b^{3}+a^{2} b^{2}\right] \\ &=(a+b)^{2}\left[\left(a^{2}\right)^{2}+(a b)^{2}+\left(b^{2}\right)^{2}-2 a^{3} b-2 a b^{3}+2 a^{2} b^{2}\right] \\ &=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)^{2} \\ &=\left(a^{3}+b^{3}\right)^{2} \\ &=R \cdot H . S \end{aligned}

Hence it is proved that

$\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$