Question

if x+1/x=4 find (x²+1/x²) and (x³+1/x³)
answer is 14,52​

Answer 1

Answer:

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Step-by-step explanation:

hope this will help....

Answer 2

Given -

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

To find -

$\bull \: {x}^{2} + \frac{1}{ {x}^{2} } = ?$

$\bull \: {x}^{3} + \frac{1}{ {x}^{3} } = ?$

Solution -

$(x + \frac{1}{x} ) = 4$

Squaring both sides,

$\implies{(x + \frac{1}{x} )}^{2} = {4}^{2}$

$\implies{x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 16$

$\implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16$

$\implies {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2$

$\implies \: {x}^\red{2} + \frac{1}{ {x}^{2} } = 14$

Again,

$\implies (x + \frac{1}{x} ) = 4$

Cubing both sides,

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

$\implies {x}^{ 3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 64$

$\implies {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 = 64$

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

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

$\implies {x}^{3} + \frac{1}{ {x}^{3} } = 52$

Formula used here -

• (a+b)² = a²+b²+2ab
• (a+b)³ = a³ +b³+3ab(a+b)