x2 + 5x + 25
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It is x = -5
Difference of two cubes: a3 - b3 = (a-b)(a2+ab+b2) 125 - x3 = 53 - x3 = (5 - x)(52 + 5x + x2) = (5 - x)(25 + 5x + x2)
x^(3) + 125 = 0 Remember that 125 = 5^(3) Hence x^(3) + 5^(3) = 0 Factor (x + 5) ( x^(2) - 5x + 25) = 0 So x = -5 & x^(2) - 5x + 125 = 0 Apply Quadratic Eq'n x = {--5 +/- sqrt[)=5)^(2) - 4(1)(125)]} / 2(1) x = { 1 +/- sqrt[25 - 500]} / 2 x = { 1 +/- sqrt[-374] } / 2 This part remain unresolved because you cannot rake the square root of a negative number. Hence x = -5 is the only result.
x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)
It is a cubic polynomial in x and its value depends on the value of x.