If x=3⅓ + 3⅔, prove that x³-9x-12 = 0 | Indices -Simplify | SciPiPupil

Question: If $x = 3^{\frac{1}{3}} + 3^{\frac{2}{3}}$, prove that: $x³-9x-12=0$.

Solution:

Given,

$x = 3^{\frac{1}{3}} + 3^{\frac{2}{3}}$

To prove: $x³-9x-12=0$

LHS:

$= x³-9x-12$

$= \left ( 3^{\frac{1}{3}} + 3^{\frac{2}{3}} \right )^{3} - 9\left ( 3^{\frac{1}{3}} + 3^{\frac{2}{3}} \right ) -12$

$= \left ( 3^{\frac{1}{3}} + 3^{\frac{2}{3}} \right )^{3} - 3²\left ( 3^{\frac{1}{3}} + 3^{\frac{2}{3}} \right ) -12$

$= \left \{ (3^{\frac{1}{3}} )^3 + 3.(3^{\frac{1}{3}})^2.3^{\frac{2}{3}} + 3.3^{\frac{1}{3}}.(3^{\frac{2}{3}} )^2 + (3^{\frac{2}{3}})^3 \right \} - 3².3^{\frac{1}{3}} - 3².3^{\frac{2}{3}} + 12$

$= \{3 + 3^{1+\frac{2}{3}+\frac{2}{3}} + 3^{1+\frac{1}{3} +\frac{4}{3}} + 3² \} - 3^{2+\frac{1}{3}} - 3^{2+\frac{2}{3}} - 12$

$= \{3 + 3^{\frac{3+2+2}{3}} + 3^{\frac{3+1+4}{3}} + 9 \} - 3^{\frac{6+1}{3}} - 3^{\frac{6+2}{8}} - 12$

$= 12 + 3^{\frac{7}{3}} + 3^{\frac{8}{3}} - 3^{\frac{7}{3}} - 3^{\frac{8}{3}} - 12$

$= 12 - 12 + 3^{\frac{7}{3}}  - 3^{\frac{7}{3}} + 3^{\frac{8}{3}} - 3^{\frac{8}{3}}$

$= 0$

RHS

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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