Question

if b^2+1/b^2= 62 than prove that b^3+1/b^3=488
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Answer 1

EXPLANATION.

⇒ (b² + 1/b²) = 62.

As we know that,

Formula of :

⇒ (x + y)³ = x³ + 3x²y + 3xy² + y³.

⇒ (x + y)² = x² + y² + 2xy.

Using this formula in the equation, we get.

⇒ (b + 1/b)² = b² + 1/b² + 2(b)(1/b).

⇒ (b + 1/b)² = b² + 1/b² + 2.

Put the values of (b² + 1/b²) = 62 in the equation, we get.

⇒ (b + 1/b)² = 62 + 2.

⇒ (b + 1/b)² = 64.

⇒ (b + 1/b) = √64.

⇒ (b + 1/b) = ± 8.

Now, cubing both sides of the equation, we get.

⇒ (b + 1/b)³ = (8)³.

⇒ (b)³ + 3(b)²(1/b) + 3(b)(1/b)² + (1/b)³ = (8)³.

⇒ b³ + 3b + 3/b + 1/b³ = 512.

⇒ b³ + 3(b + 1/b) + 1/b³ = 512.

Put the values of (b + 1/b) = 8 in the equation, we get.

⇒ b³ + 3(8) + 1/b³ = 512.

⇒ b³ + 1/b³ + 24 = 512.

⇒ b³ + 1/b³ = 512 - 24.

⇒ b³ + 1/b³ = 488.

Hence Proved.