Question

If X =
${3}^{ \frac{1}{3} } - {3}^{ - \frac{ 1}{3} }$
Show that :

$3 {x}^{3} + 9x \: = 8$
​

Answer 1

Answer:

$\rm{heya \: mate} \\ \\ \rm{ \: refers \: in \: the \: attahment}$

Answer 2

Given :-

• x = 3^(1/3) - 3^(-1/3)

To Prove :-

• 3x³ + 9x = 8 ?

Solution :-

→ x = 3^(1/3) - 3^(-1/3)

Cubing both sides we get,

→ x³ = [ 3^(1/3) - 3^(-1/3) ]³

Now, using (a - b)³ = a³ - b³ - 3ab(a - b) in RHS , we get,

→ x³ = {3^(1/3)}³ - {3^(-1/3)}³ - 3 * 3^(1/3) * 3^(-1/3) [ 3^(1/3) - 3^(-1/3) ]

Using (a^b)^c = (a)^(bc) and, a^b * a^c = a^(b + c)

→ x³ = (3)^(1/3 * 3) - (3)^(-1/3 * 3) - 3 * 3^( -1/3 + 1/3) * [ 3^(1/3) - 3^(-1/3) ]

→ x³ = 3 - 3^(-1) - 3 * 3^0 * [ 3^(1/3) - 3^(-1/3) ]

using a^(-b) = 1/a^b and a^0 = 1 , and putting [ 3^(1/3) - 3^(-1/3) ] = x

→ x³ = 3 - 1/3 - 3 ( x )

→ x³ = [ 3 - 1/3 - 3x]

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Putting value of x³ now, we get :-

→ 3x³ + 9x

→ 3[3 - 1/3 - 3x] + 9x

→ 9 - 1 - 9x + 9x

→ 8 (Hence, Proved).

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