Question

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Answer 1

Question :-

If x = 1/(4 - √15), y = 1/(4 + √15), then find the value of x³ + y³ is

(A) 486

(B) 439

(C) 488

(D) 476

Solution :-

x = 1/(4 - √15)

y = 1/(4 + √15)

Finding the value of x + y

$\sf x + y = \dfrac{1}{4 - \sqrt{15} } + \dfrac{1}{4 + \sqrt{15} } \\ \\ \\ \tt \underline{taking \ lcm} \\ \\ \\ \sf = \dfrac{4 + \sqrt{15} + 4 - \sqrt{15} }{(4 - \sqrt{15})(4 + \sqrt{15)} } \\ \\ \\ \sf = \dfrac{8}{ {4}^{2} - {( \sqrt{15}) }^{2} } \qquad \bigg[ \because (x - y)(x + y) = {x}^{2} - {y}^{2} \bigg] \\ \\ \\ \sf = \dfrac{8}{16 - 15} \\ \\ \\ \sf = \dfrac{8}{1} \\ \\ \\ \rm x + y = 8$

Finding the value of xy

$\sf xy = \bigg( \dfrac{1}{4 - \sqrt{15} } \bigg) \bigg( \dfrac{1}{4 + \sqrt{15} } \bigg) \\ \\ \\ \sf = \dfrac{1}{(4 - \sqrt{15})(4 + \sqrt{15})} \\ \\ \\ \sf = \dfrac{1}{ {4}^{2} - {( \sqrt{15})}^{2} } \qquad \bigg[ \because (x - y)(x + y) = {x}^{2} - {y}^{2} \bigg] \\ \\ \\ \\ \sf = \dfrac{1}{16 - 15} \\ \\ \\ \sf = \dfrac{1}{1} \\ \\ \\ \rm xy = 1$

We know that

(x + y)³ = x³ + y³ + 3xy(x + y)

Here

• x + y = 8

• xy = 1

By substituting the values in the above identity

⇒ (8)³ = x³ + y³ + 3(1)(8)

⇒ 512 = x³ + y³ + 24

⇒ 512 - 24 = x³ + y³

⇒ 488 = x³ + y³

⇒ x³ + y³ = 488

Therefore the value of x³ + y³ is 488 i.e option (C).

Answer 2

Your answer is option C