Well explained thanks👍

SENSATIONAL!!

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Easy way to solve this equation 1. Cube both side 27X^3 = ((3X^2 - 12)^2)^(1/3) + 108X 2. Simplify the equation to get X^4 - 3X^3 - 8X^2 +12X + 16 = 0 (X^4 -8X^2 +16) - 3X^3 +12 = 0 (X^2 - 4)^2. - 3X(X^2 -4) =0 (X^2 -4)(X^2 - 4 - 3X) = 0 Hence X^2 - 4 = 0 or X^2 -3X -4 =0 Solving these equation we get X = 2., -2, -1 and 4 Thanks

At 2:58 you have arrived at the quartic equation x⁴ − 3x³ − 8x² + 12x + 16 = 0 and then you spend the next _seven_ minutes solving this equation, but your solution is unnecessarily longwinded and complicated. Note that x⁴ − 8x² + 16 = (x² − 4)² is a perfect square and that we can factor out 3x from the two remaining terms so we can write the equation as (x² − 4)² − 3x(x² − 4) = 0 and taking out the now common factor (x² − 4) this gives (x² − 4)(x² − 4 − 3x) = 0 (x + 2)(x − 2)(x + 1)(x − 4) = 0 x = −2 ⋁ x = 2 ⋁ x = −1 ⋁ x = 4

X=2

X=2