The factors of $x^{4} + 64$ are:

(x^{2} + 8)^{2}

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(x^{2} + 8) (x^{2} - 8)

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(x^{2} + 2 x + 4) (x^{2} - 8 x + 16)

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(x^{2} - 4 x + 8) (x^{2} + 4 x + 8)

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Solution

The correct option is B $(x^{2} - 4 x + 8) (x^{2} + 4 x + 8)$
$x^{4} + 64 = x^{4} + 64 + 16 x^{2} - 16 x^{2} = {(x^{2} + 8)}^{2} - 16 x^{2} = (x^{2} + 4 x + 8) (x^{2} - 4 x + 8) [∵ a^{2} - b^{2} = (a + b) (a - b)]$

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The expression $x^{4}$ + 8 $x^{2}$ + 16 can be factorized as

The expression $x^{4} + 8 x^{2} + 16$ can be factorized as:

Factorise: x^{4} - (x - 4)^{4}

Find the third side of a triangle whose two sides are $(3 x^{2} - 5 x)$ cm and $(4 x^{2} + 12)$ cm and perimeter is $(8 x^{2} - 9 x + 4)$ cm.

\frac{x^{2} - 16}{x^{2} - 3 x + 2} \times \frac{x^{2} - 4}{x^{3} + 64} \times \frac{x^{2} - 4 x + 16}{x^{2} - 2 x - 8}

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