# How do you solve (t+1)^3=t^3+7?

Jun 9, 2016

$t = 1$ or $t = - 2$

#### Explanation:

Expanding ${\left(t + 1\right)}^{3} = {t}^{3} + 7$ using identity ${\left(t + 1\right)}^{3} = {t}^{3} + 3 {t}^{2} + 3 t + 1$ above can be written as

${t}^{3} + 3 {t}^{2} + 3 t + 1 = {t}^{3} + 7$ or

$3 {t}^{2} + 3 t - 6 = 0$ or dividing by $3$

${t}^{2} + t - 2 = 0$

Splitting middle term $t$ as $2 t - t$, above is equivalent to

${t}^{2} + 2 t - t - 2 = 0$ or

$t \left(t + 2\right) - 1 \left(t + 2\right) = 0$ or

$\left(t - 1\right) \left(t + 2\right) = 0$ or

$t = 1$ or $t = - 2$

Jun 9, 2016

The two solutions are $- 2$ and $1$

#### Explanation:

Expand the cube, using the formula

 (a+b)^3 = a^3+3a^2b+3ab^2+b^3

In words, the cube of a sum is given by the sum of the two cubes, and the triple product of the square of an element and the other. In our case:

 (t+1)^3 = t^3+3t^2*1+3t*1^2+1^3

= t^3+3t^2+3t+1

Writing the whole equation, we see that the cubes cancel out:

 cancel(t^3)+3t^2+3t+1 = cancel(t^3)+7

And subtracting $7$ from both members, we have

 3t^2+3t-6=0 iff t^2+t-2=0 iff (t+2)(t-1)=0

Thus, the two solutions are $t = - 2$ and $t = 1$.