Question

If (x+1/x)²=3
Prove that (x³+1/x³)=0

Answer 1

ANSWER

WE KNOW

(X+1/X) ^3 = X3+1/X3 +3(X+1/X)

SO

X3+1/X3 = (x+1/x) ^3 -3(x+1/x)

= 3√3-3√3

= 0

{since

(x+1/x) = √3

which implies (x+1/x) ^3 = 3√3}

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Answer 2

Step-by-step explanation:

Given : ${\sf{\ \ \left( x + {\dfrac{1}{x}} \right) ^2 = 3}}$

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${\boxed{\tt{\bigstar \ Identity \ : \ (a + b)^2 = a^2 + b^2 + 2ab}}}$

${\tt{Here, \ a = x, \ b = {\dfrac{1}{x}} }}$

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On further solving, we get

$\implies{\sf{ (x)^2 + \left( {\dfrac{1}{x}} \right) ^2 + 2 . x . {\dfrac{1}{x}} = 3}}$

$\implies{\sf{ x^2 + {\dfrac{1}{x^2}} + 2 = 3}}$

$\implies{\sf{ x^2 + {\dfrac{1}{x^2}} = 3 - 2}}$

$\implies{\sf{ x^2 + {\dfrac{1}{x^2}} = 1 \qquad \qquad ...(1)}}$

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To Prove : ${\sf{\ \ x^3 + {\dfrac{1}{x^3}} = 0}}$

_____________________________

${\boxed{\tt{\bigstar \ Identity \ : \ a^3 + b^3 = (a + b)(a^2 + b^2 - ab)}}}$

${\tt{Here, \ a = x, \ b = {\dfrac{1}{x}} }}$

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L.H.S. = ${\sf{ x^3 + {\dfrac{1}{x^3}} }}$

$\implies{\sf{ \left( x + {\dfrac{1}{x}} \right) \left[ (x)^2 + \left( {\dfrac{1}{x}} \right) ^2 - x . {\dfrac{1}{x}} \right] }}$

$\implies{\sf{ \left(x + {\dfrac{1}{x}} \right) \left( x^2 + {\dfrac{1}{x^2}} - 1\right) }}$

$\implies{\sf{ \left(x + {\dfrac{1}{x}} \right) (1 - 1) \qquad ...[ From (1) ] }}$

$\implies{\sf{ \left(x + {\dfrac{1}{x}} \right) (0) }}$

$\implies{\sf{ 0 }}$

= R.H.S.

Hence, proved !!