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in Linear Inequations by (46.2k points)
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If (a2 + b2)3 = (a3+b3)2 and ab ≠ 0, then \(\big(\frac{a}{b}+\frac{b}{a}\big)^6\) is equal to

(a) \(\frac{a^6 + b^6}{a^3b^3}\)

(b) \(\frac{64}{729}\)

(c) 1

(d) \(\frac{a^6 + a^3b^3 + b^6}{a^2b^4+a^4b^2}\)

1 Answer

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Best answer

(b) \(\frac{64}{729}.\)

(a2 + b2)3 = (a3+b3)2

a6 + b6 + 3a2b2 (a2 + b2) = a6 + b6 + 2a2b2

⇒ a2 + b\(\frac23\) ab             .....(i)

Now, \(\big(\frac{a}{b}+\frac{b}{a}\big)^6\) = \(\big(\frac{a^2+b^2}{ab}\big)^6\) = \(\bigg(\frac{\frac23ab}{ab}\bigg)^6\)\(\big(\frac23\big)^6\) = \(\frac{64}{729}.\)

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