Question
Evaluate the following using identities:$\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2$
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Answer
We have,$\big(\text{a}^2\text{b}-\text{a}\text{b}^2\big)^2$
$=\big(\text{a}^2\text{b}\big)^2+\big(\text{ab}^2\big)^2-2\times\text{a}^2\times\text{ab}^2$ $\begin{bmatrix}\because\big(\text{a}-\text{b}\big)^2=\text{a}^2+\text{b}^2-2\text{ab}\\\text{Where}\ \text{a}=\text{a}^2\text{b}\ \text{and}\ \text{b}=\text{ab}^2\end{bmatrix}$
$=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
$\therefore\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
$=\big(\text{a}^2\text{b}\big)^2+\big(\text{ab}^2\big)^2-2\times\text{a}^2\times\text{ab}^2$ $\begin{bmatrix}\because\big(\text{a}-\text{b}\big)^2=\text{a}^2+\text{b}^2-2\text{ab}\\\text{Where}\ \text{a}=\text{a}^2\text{b}\ \text{and}\ \text{b}=\text{ab}^2\end{bmatrix}$
$=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
$\therefore\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
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