If (x+1x)2=3, find x3+1x3.\Big(x + \dfrac{1}{x}\Big)^2 = 3, \text{ find } x^3 + \dfrac{1}{x^3}.(x+x1)2=3, find x3+x31.
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Given,
⇒(x+1x)2=3∴x+1x=±3.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3 \\[1em] \therefore x + \dfrac{1}{x} = \pm \sqrt{3}.⇒(x+x1)2=3∴x+x1=±3.
We know that,
x3+1x3=(x+1x)3−3(x+1x)x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big)x3+x31=(x+x1)3−3(x+x1)
When x+1x=3x + \dfrac{1}{x} = \sqrt 3x+x1=3 substituting values we get,
x3+1x3=(3)3−3×3=33−33=0.x^3 + \dfrac{1}{x^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0.x3+x31=(3)3−3×3=33−33=0.
When x+1x=−3x + \dfrac{1}{x} = -\sqrt{3}x+x1=−3 substituting values we get,
x3+1x3=(−3)3−3×(−3)=−33+33=0.x^3 + \dfrac{1}{x^3} = (-\sqrt{3})^3 - 3 \times (-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0.x3+x31=(−3)3−3×(−3)=−33+33=0.
Hence, x3+1x3=0x^3 + \dfrac{1}{x^3} = 0x3+x31=0.
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