Question

(3/2)^-1 × (3/2)^-1 × (3/2)^-1​

Answer 1

Answer:

so x - 2 = 2^(2/3) + 2^(1/3)

=(x-2)^3 = [2^(2/3)+2^(1/3)]^3

use (a-b)^3= a^3 - b^3 - 3ab(a-b)

and (a+b)^3= a^3 + b^3 + 3ab(a+b) formulae.

or x^3 - 8 - 6x(x-2) = 2^2 + 2^1 + 3*[2^{(2/3)+(1/3)}[2^(2/3)+2^(1/3)

or x^3 - 8 - 6x^2 + 12x = 4 + 2 + 6(x-2)

or x^3 - 8 -6x^2 + 12x = 6 + 6x - 12

or x^3 - 6x^2 +6x = 2

Answer 2

$\green{\bold{\underline{ ☆ UPSC-ASPIRANT ☆} }}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

(3/2)^-1 × (3/2)^-1 × (3/2)^-1

$\huge\bigstar\huge\tt\underline\red{ᴀɴsᴡᴇʀ }\bigstar$

$➜ { (\frac{3}{2}) }^{ - 1} \times { (\frac{3}{2} )}^{ - 1} \times {( \frac{3}{2} )}^{ - 1}$

$➜ { (\frac{2}{3}) }^{1} \times { (\frac{2}{3} )}^{1} \times { (\frac{2}{3} )}^{1}$

$➜ \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$

Here is your answer^^^^

Hope it helps you..!!!

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Thankyou:)