Question

please help in following question​

Answer 1

Answer:

Option (c) is correct

Step-by-step explanation:

$i) x = \frac{1}{4-\sqrt{15}}$

$\implies x = \frac{4+\sqrt{15}}{(4-\sqrt{15})(4+\sqrt{15})}$

$= \frac{4+\sqrt{15}}{4^{2}-(\sqrt{15})^{2}}$

$= \frac{4+\sqrt{15}}{16-15}$

$= 4+\sqrt{15}\: ---(1)$

$ii) y = \frac{1}{4+\sqrt{15}}$

$\implies y= \frac{4-\sqrt{15}}{(4+\sqrt{15})(4-\sqrt{15})}$

$= \frac{4-\sqrt{15}}{4^{2}-(\sqrt{15})^{2}}$

$= \frac{4-\sqrt{15}}{16-15}$

$= 4-\sqrt{15}\: ---(2)$

$iii) x+y \\= 4+\sqrt{15}+4-\sqrt{15}\\=8\:--(3)$

$xy =( 4+\sqrt{15})(4-\sqrt{15})\\=4^{2}-(\sqrt{15})^{2}\\=16-15\\ = 1\:---(4)$

$Now, x^{3}+y^{3}\\=(x+y)^{3}-3xy(x+y)\\=8^{3}-3\times 1\times 8\\=8(8^{2}-3)\\=8(64-3)\\=8\times 61\\=488$

Therefore,

Option (C) is correct

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