Question

if a square + 1/a square =47 and a ≠0 find a cube +1 / a cube​

Answer 1

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Answer 2

Answer:

${a}^{2} + \frac{1}{ {a}^{2} } = 47$

$\Rightarrow$ To find

${a}^{3} + \frac{1}{ {a}^{3} }$

${(a + \frac{1}{a} )}^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2$

Simplified using the identity (a+b)^2.

${(a + \frac{1}{a} })^{2} = 47 + 2 = 49 \\ a + \frac{1}{a} = \sqrt{49} = 7$

Using the identity a^3 + b^3 = (a+b)(a^2-ab+b^2)

${a}^{3} + \frac{1}{ {a}^{3} } = (a + \frac{1}{a} )( {a}^{2} + \frac{1}{ {a}^{2} } - 1)$

=$7(47 - 1) = 7 \times 46 = 322$