Question

If X=(ab)^1/3-(ab)^(-1/3), prove that x^3+1/ab+3x=ab

Answer 1

Step-by-step explanation:

x = $(ab)^{\frac{1}{3} }$ - $(ab)^{\frac{-1}{3} }$ ...(i)

Cubing both sides,

x³ = ($(ab)^{\frac{1}{3} }$ - $(ab)^{\frac{-1}{3} }$)³

x³ = ($(ab)^{\frac{1}{3} }$)³ - ($(ab)^{\frac{-1}{3} }$)³ - 3 . $(ab)^{\frac{1}{3} }$.$(ab)^{\frac{-1}{3} }$ . ($(ab)^{\frac{1}{3} }$ - $(ab)^{\frac{-1}{3} }$)

x³ = ab - $(ab)^{-1}$ - 3 . $(ab)^{\frac{1}{3}-\frac{1}{3} }$ . x

x³ = ab - $\frac{1}{ab}$ - 3.1.x

x³ + $\frac{1}{ab}$ +3x = ab

Proved

Answer 2

Answer:

Step-by-step explanation:

Taking LHS,

= x^3+1/ab+3x

= {(ab)^1/3 - 1/(ab)^1/3}^3 + 3{(ab)^1/3 - 1/(ab)^1/3} + 1/ab

= (ab)^1/3*3 - 1/(ab)^1/3*3 - 3 {(ab)^ 1/3 - 1/(ab)^1/3 } + 3x +1/(ab)

= ab - 1/ab -3x +3x +1/(ab)

= ab

= proved