Let $x$ be real number such that $x^{3} + 4 x = 8,$ then the value of $x^{7} + 64 x^{2}$ is.

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Solution

$x ϵ R$
$x^{3} + 4 x = 8$ Multiply with both side $x^{4}$
$\Rightarrow x^{7} + 4 x^{5} = 8 x^{4}$ ___ (i)
$x^{3} + 4 x = 8$ Multiply with $x^{2}$ both side
$\Rightarrow x^{5} + 4 x^{3} = 8 x^{2}$ (ii)
$e q (i) - 4 \times e q (i i)$
$\Rightarrow x^{7} - 16 x^{3} = 8 x^{4} - 32 x^{2}$
$x^{7} = - 32 x^{2} + 16 x^{3} + 8 x^{4}$
$= - 32 x^{2} + 16 (8 - 4 x) + 8 x^{4}$
$= - 32 x^{2} + 128 - 64 x + 8 x^{4}$
$= - 32 x^{2} + 128 - 8 x (8 - x^{3})$
$x^{7} = - 32 x^{2} + 128 - 8 x (4 x)$
$x^{7} = - 64 x^{2} + 128$
$x^{7} + 64 x^{2} = 128$

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