1.

If \(x - \frac 3 x = 6,\; x \ne 0,\) then the value of \(\frac {x^4 - \frac {27}{x^2}}{x^2 - 3x - 3}\) is:1. 542. 2703. 804. 90

Answer» Correct Answer - Option 4 : 90

GIVEN:

\(x - \frac 3 x = 6,\; x \ne 0,\)

FORMULA USED:

(a - b)2 = (a2 + b2 - 2ab), (a3 - b3) = (a - b) (a2 + b2 + ab)

CALCULATION:

(a - b)2 = (a2 + b2 - 2ab)

⇒ \(x - \frac 3 x = 6,\; x \ne 0,\)

Squaring both sides

⇒ (x - 3/x)2 = (6)2

⇒ x2 + 9/x2 - 6 = 36

⇒ x2 + 9/x2 = 42

\(\frac {x^4 - \frac {27}{x^2}}{x^2 - 3x - 3}\)

 Divide by x into numerator and denominator

\(⇒ \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - 3 - \frac{3}{x}}}\)

\( ⇒ \frac{{\left( {x - \frac{3}{x}} \right)\left( {{x^2} + \frac{9}{{{x^2}}} + 3} \right)}}{{\left( {x - \frac{3}{x}} \right) - 3}}\)

\(⇒ \;\frac{{6 \times \left( {42 + 3} \right)}}{{6 - 3}}\)

\( ⇒ \;\frac{{6 \times 45}}{3}\)

⇒ 90

∴ \( \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - 3 - \frac{3}{x}}}\) = 90


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