Question
Find the product $a b\left(a^2+b^2\right)$ and evaluate it fore a = 2 and $\text{b}=\frac{1}{2}$.
✓
Answer
$a b\left(a^2+b^2\right)$
$=a b \times a^2+a b \times b^2$
$=a \times a^2 \times b+a \times b \times b^2$
$=a^{(1+2)} \times b+a \times b^{(1+2)}$
$=a^3 b+a b^3$
When $\mathrm{a}=2$ and $\mathrm{b}=\frac{1}{2}$, we get,
$\text { L.H.S }=a b\left(a^2+b^2\right)$
$=2 \times \frac{1}{2}\left(2^2+\frac{2}{2^2}\right)$
$=4+\frac{1}{4}=\frac{17}{4}$
$\text { R.H.S }=a^3 b+a^3$
$=2^3 \times \frac{1}{2}+2\left(\frac{1}{2}\right)^3$
$=4+\frac{1}{4}=\frac{17}{4}$
$\therefore$ $L.H.S.$ $=$ $R.H.S.$
$=a b \times a^2+a b \times b^2$
$=a \times a^2 \times b+a \times b \times b^2$
$=a^{(1+2)} \times b+a \times b^{(1+2)}$
$=a^3 b+a b^3$
When $\mathrm{a}=2$ and $\mathrm{b}=\frac{1}{2}$, we get,
$\text { L.H.S }=a b\left(a^2+b^2\right)$
$=2 \times \frac{1}{2}\left(2^2+\frac{2}{2^2}\right)$
$=4+\frac{1}{4}=\frac{17}{4}$
$\text { R.H.S }=a^3 b+a^3$
$=2^3 \times \frac{1}{2}+2\left(\frac{1}{2}\right)^3$
$=4+\frac{1}{4}=\frac{17}{4}$
$\therefore$ $L.H.S.$ $=$ $R.H.S.$
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