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Answer 1 · 2018-02-07T08:47:38

$x=4$

By inspection you can see that $(4)^3-(4)^2=64-16=48$ .

If we rearrange we get $x^3-x^2-48=0$ .

Knowing that $x=4$ is a zero means that $x-4$ is a factor.

Dividing we have:

$(x^3-x^2-48)/(x-4) = x^2+3x+12$

So we can say that:

$x^3-x^2-48 =(x-4)(x^2+3x+12)$

so

$x^3-x^2-48=0$

is equivalent to

$(x-4)(x^2+3x+12)=0$

We already known $x=4$ is a solution.

$x^2+3x+12=0$ has no real solutions since the discriminant is less than 0: $9-4(12)<0$ .

The only solution is $x=4$ , which I found by inspection, but we could also find by the rational root theorem.

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