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SSC CHSL 2021 PYQ — Question 62

SSC CHSL - 19 April 2021 - Shift 2 • Quantitative Aptitude • Q62

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If x² - 5√2x - 1 = 0, then what will be the value of x3−1x2x^3-\frac{1}{x^2}

 

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Correct answer: D

Explanation

The correct answer is option 2 i.e. 265√2

Given:

x² - 5√2x - 1 = 0

Fomula Used:

(a - b)3 = a3 - b3 - 3ab(a - b)

Explanation:

According to the gievn question,

x² - 5√2x - 1 = 0

Dividing the bove equation by x, we get

⇒ x−1x=52x - {1 \over x} = 5\sqrt{2}

Now taking cube on both sides, we get

 ⇒ (x−1x)3=(52)3(x - {1 \over x})^3 = (5\sqrt{2})^3

 ⇒ x3−1x3−3(x)(1x)(x−1x)=125×22x^3 - {1 \over x^3} - 3(x)({1\over x})(x - {1 \over x})= 125 \times 2\sqrt{2}

 ⇒ x3−1x3−3(52)=2502x^3 - {1 \over x^3} - 3(5 \sqrt{2})= 250\sqrt{2}

⇒ x3−1x3−152=2502x^3 - {1 \over x^3} - 15 \sqrt{2}= 250\sqrt{2}

⇒ x3−1x3=2502+152x^3 - {1 \over x^3} = 250\sqrt{2} + 15 \sqrt{2}

⇒ x3−1x3=2652x^3 - {1 \over x^3} = 265\sqrt{2}

Hence the required answer is 265√2.