Identities

x^4+x^4 + 2x^2(1/x^2) = 322+2
(x^2+(1/x^2))^2 = 324
x^2+(1/x^2)= 18
x^2+(1/x^2)- 2x(1/x) = 18-2
x-(1/x)=4
Now cubing both sides
x^3 - (1/x^3) - 3x(1/x)(x-(1/x)) = 64
x^3-(1/x^3) - 3.4 =64
x^3-(1/x^3) = 64 + 12 = 76
And there you have it as b

x⁴ + 1x⁴ = 322
or, (x²)² + (1x²)² + 2.x².1x²
= 322 + 2 = 324 = 18²

or, (x² + 1x²)² = 18²

or, x² + 1x² = 18

or, x² + (1x)² - 2.x.1x
= 18-2 = 16 = 4²

Continuing with the sum,

(x - 1x)² = 4²
or, x - 1x = 4
or, (x - 1x)³ = 4³
or, x³ - 3.x².1x + 3.x.(1x)² - 1x³ = 64
Therefore, x³ - 1x³ - 3x + 3x = 64
Therefore, x³ - 1x³ -3(x - 1x) = 64
Therefore,
x³ - 1x³ - 3 * 4 = 64
Or, x³ - 1x³ - 12 = 64

Or, x³ - 1x³ = 64 + 12 = 76

Ans) x³ - 1x³ = 64 + 12 = 76

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