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