If $a + b = 10$ and $a b = 16$ find, the value of $a^{2} - a b + b^{2}$ and $a^{2} + a b + b^{2}$

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Solution

Given : $a + b = 10$ ............(1)
and $a b = 16$ .....................(2)
It is known that,
$(a + b)^{2} = a^{2} + 2 a b + b^{2}$
Now, substituting the values of (1) and (2) in the above, we get.
$\Rightarrow (10)^{2} = a^{2} + 2 \times (16) + b^{2}$
$\Rightarrow 100 = a^{2} + 32 + b^{2}$
$\Rightarrow a^{2} + b^{2} = 100 - 32$
$\Rightarrow a^{2} + b^{2} = 68$ .................(3)
Now, substituting the value of $a^{2} + b^{2} = 68$ in $a^{2} - a b + b^{2}$
$= 68 - 16$ (as $a b = 16$ )
$= 52$
And, substituting the value of $a^{2} + b^{2} = 68$ in $a^{2} + a b + b^{2}$
$= 68 + 16$
$= 84$

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