Question

if x2 + 1/x2=98 find the value of x3+1/x3

Answer 1

hey! bro..
#your Ans
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$(x + \frac{1}{x} {)}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ \\ (x + \frac{1}{x} {)}^{2} = 98 + 2 \\ \\ (x + \frac{1}{x} {)}^{2} = 100 \\ \\ x + \frac{1}{x} = \sqrt{100} \\ \\ x + \frac{1}{x} = 10 \\ \\ now \\ \\ (x + \frac{1}{x} {)}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) \\ \\ putting \: value \\ \\ (10 {)}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 10 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 1000 - 30 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 970 \: ans$

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Answer 2

Given:

$x^2+\dfrac{1}{x^2}=98$

To find:

The value of $x^3+\dfrac{1}{x^3}$

Solution:

Formulae used:

$(a+b)^2=a^2+2ab+b^2$

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

We know that,

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

$(x+\dfrac{1}{x})^2=98+2(1)$           $[\because x^2+\dfrac{1}{x^2}=98]$

$(x+\dfrac{1}{x})^2=98+2$

$(x+\dfrac{1}{x})^2=100$

Taking square root on both sides, we get

$x+\dfrac{1}{x}=10$         ...(i)

Now,

$(x+\dfrac{1}{x})^3=x^3+\dfrac{1}{x^3}+3\times x\times \dfrac{1}{x}\left(x+\dfrac{1}{x}\right)$

$(10)^3=x^3+\dfrac{1}{x^3}+3\left(10\right)$        [Using (i)]

$1000=x^3+\dfrac{1}{x^3}+30$

Subtract 30 form both sides.

$1000-30=x^3+\dfrac{1}{x^3}$

$970=x^3+\dfrac{1}{x^3}$

Therefore, the value of $x^3+\dfrac{1}{x^3}$ is 970.