Question

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If x^2 + 1/x^2=14, find the value ofx^3 + 1/x^3

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\sf{{x}^{3} + \dfrac{1}{ {x}^{3}} = 52}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{{x}^{2} + \dfrac{1}{ {x}^{2} } = 14}$

To find :- $\sf{{x}^{3} + \dfrac{1}{ {x}^{3} }}$

Solution :-

${x}^{2} + \dfrac{1}{ {x}^{2} } = 14$

Now add 2 on both sides

${x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 14 + 2$

${x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16$

Now it can be written as,

${x}^{2} + \dfrac{1}{ {x}^{2} } + 2(x)( \dfrac{1}{x}) = 16$

We know that, (x + y)² = x² + y² + 2xy

Here x = x, y = 1/x

By substituting the values in the identity we have,

$(x + \dfrac{1}{x})^{2} = 16$

$x + \dfrac{1}{x} = \sqrt{16}$

$x + \dfrac{1}{x} = 4$

Now, $\boxed{\sf{x + \dfrac{1}{x} = 4}}$

By cubing on both sides,

${(x + \dfrac{1}{x})}^{3} = {4}^{3}$

${(x + \dfrac{1}{x})}^{3} = 64$

We know that, (x + y)³ = x³ + y³ + 3xy(x + y)

Here x = x, y = 1/x

By substituting the values in the identity we have,

${x}^{3} + \dfrac{1}{ {x}^{3} } + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 64$

${x}^{3} + \dfrac{1}{ {x}^{3} } + 3(x + \dfrac{1}{x}) = 64$

${x}^{3} + \dfrac{1}{ {x}^{3} } + 3(4) = 64$

[Since $\bf{x + \dfrac{1}{x} = 4}$ ]

${x}^{3} + \dfrac{1}{ {x}^{3} } + 12 = 64$

${x}^{3} + \dfrac{1}{ {x}^{3} } = 64 - 12$

${x}^{3} + \dfrac{1}{ {x}^{3} } = 52$

$\Huge{\boxed{\sf{{x}^{3} + \dfrac{1}{ {x}^{3}} = 52}}}$

$\mathfrak{\large{\underline{\underline{Identity \: used:-}}}}$

1. (x + y)² = x² + y² + 2xy

2. (x + y)³ = x³ + y³ + 3xy(x + y)

Answer 2

ANSWER:

${x}^{2} + \dfrac{1}{x^{2} } = 14$

REQUIRED TO FIND:

we need to find the value of

${x}^{3} + \dfrac{1}{ {x}^{3} }$

METHOD:

in the attachment

first they have given us the value of

$x^{2} + \dfrac{1}{x ^{2} } = 14$

we are going to add 2 on both sides

so that this will become of the form

${(a + b)}^{2}$

so when we solve we get two values of

$x + \dfrac{1}{x}$

so we should substitute both the values or just take 4

IDENTITIES USED:

$> {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

${(a + b}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$