Best Answer

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.

Q: How do you solve x3 plus 125 is equals to 0?

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It is x = -5

x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)

IMPOSSIBLE

x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.

It is a cubic polynomial in x and its value depends on the value of x.

Related questions

It is x = -5

x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)

IMPOSSIBLE

cube root of 216 = 6 > X= 6

x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.

It is a cubic polynomial in x and its value depends on the value of x.

No.

y2=x3+3x2

No, it is not.

(xn+2-1)/(x2-1)

x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)

x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4