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Question: How do you solve \({x^3} + 64\)=0?How do you solve \({x^3} + 64\)=0?...

How do you solve x3+64{x^3} + 64=0?How do you solve x3+64{x^3} + 64=0?

Explanation

Solution

The other two special factoring formulas you will need to memorize are very similar to one another, they are the formulas for factoring the sums and the difference of cubes.

Here are the two formulas:

Factoring a sum of cubes:
a3+b3=(a+b)(a2−ab+b2){a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})

Factoring a difference of cubes:
a3−b3=(a−b)(a2+ab+b2){a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})

Complete Step-by-step solution:
Notice that this is a sum of cubes, which is factorable as follows:
a3+b3=(a+b)(a2−ab+b2){a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})

Thus, x3+64{x^3} + 64is factorable into-
x3+43=(x+4)(x2−4x+16)=0{x^3} + {4^3} = (x + 4)({x^2} - 4x + 16) = 0

Now we have one linear factor and one quadratic factor.
(x+4)(x2−4x+16)=0(x + 4)({x^2} - 4x + 16) = 0

We can set each of these equal to 00individually to find the values of xxthat make the whole expression equal 00
x+4=0⇒x=−4x + 4 = 0 \Rightarrow x = - 4

The next requires the quadratic formula.
x2−4x+16=0⇒x =4±16−642 ⇒x=4±432⇒x=2±23  {x^2} - 4x + 16 = 0 \Rightarrow x \\\ = \dfrac{{4 \pm \sqrt {16 - 64} }}{2} \\\ \Rightarrow x = \dfrac{{4 \pm 4\sqrt 3 }}{2} \Rightarrow x = 2 \pm 2\sqrt 3 \\\

These are two imaginary solutions.

Additional Information:
A zero of a function is an interception between the function itself and the x-axis.

No zero (e.g., y=x2+1)y = {x^2} + 1)graph \left\\{ {{x^2} + 1\left[ { - 10,10, - 5,5} \right]} \right\\}

One zero (e.g.y=x)(e.g.y = x) graph \left\\{ {x\left[ { - 10,10, - 5,5} \right]} \right\\}

Two or more zeros (e.g.y=x2−1)(e.g.y = {x^2} - 1) and \left\\{ {{x^2} - 1\left[ { - 10,10, - 5,5} \right]} \right\\}

Infinite zeros (e.g.y=sin⁡x)(e.g.y = \sin x)graph \left\\{ {\sin x\left[ { - 10,10, - 5,5} \right]} \right\\}
To find the eventual zeros of a function it is necessary to solve the equation system between the equation of the function and the equation of the x−axis(y=0)x - axis(y = 0)

Note: To help with the memorization, first notice that the terms in each of the two factorization formulas are exactly the same. Then notice that each formula has only one minus sign. The distinction between the two formulas is in the location of that one MINUS sign- For the difference of cubes, the MINUS sign goes in the linear factor, a−ba - b; for the sum of cubes, the MINUS sign goes in the quadratic factor goes like the following equation:
a2−ab+b2{a^2} - ab + {b^2}