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Mathematics

If x2+1x2=27x^2 + \dfrac{1}{x^2} = 27, find the value of 3x3+5x−3x3−5x.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x}.

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Answer

We know that,

⇒(x−1x)2=x2+1x2−2⇒x−1x=x2+1x2−2.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{x^2 + \dfrac{1}{x^2} - 2}.

Substituting values we get,

x−1x=27−2=25=±5.x - \dfrac{1}{x} = \sqrt{27 - 2} = \sqrt{25} = \pm 5.

Here, 3x3+5x−3x3−5x3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} can be written as,

3x3+5x−3x3−5x=3(x3−1x3)+5(x−1x)=3[(x−1x)3+3(x−1x)]+5(x−1x).3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3\Big(x^3 - \dfrac{1}{x^3}\Big) + 5\Big(x - \dfrac{1}{x}\Big) \\[1em] = 3\Big[\Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big)\Big] + 5\Big(x - \dfrac{1}{x}\Big).

Considering x−1x=5x - \dfrac{1}{x} = 5 in first case and substituting value we get,

3x3+5x−3x3−5x=3(53+3×5)+(5×5)=3(125+15)+25=(3×140)+25=420+25=445.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3(5^3 + 3 \times 5) + (5 \times 5) \\[1em] = 3(125 + 15) + 25 \\[1em] = (3 \times 140) + 25 \\[1em] = 420 + 25 \\[1em] = 445.

Considering x−1x=−5x - \dfrac{1}{x} = -5 in second case and substituting value we get,

3x3+5x−3x3−5x=3[(−5)3+3×(−5)]+[5×(−5)]=3(−125−15)−25=(3×−140)−25=−420−25=−445.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3[(-5)^3 + 3 \times (-5)] + [5 \times (-5)] \\[1em] = 3(-125 - 15) - 25 \\[1em] = (3 \times -140) - 25 \\[1em] = -420 - 25 \\[1em] = -445.

Hence, 3x3+5x−3x3−5x=±445.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = \pm 445.

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