Question

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

Answer 1

`t=1` or `t=-2`

Explanation:

Expanding `(t+1)^3=t^3+7` using identity `(t+1)^3=t^3+3t^2+3t+1` above can be written as

`t^3+3t^2+3t+1=t^3+7` or

`3t^2+3t-6=0` or dividing by `3`

`t^2+t-2=0`

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

`t^2+2t-t-2=0` or

`t(t+2)-1(t+2)=0` or

`(t-1)(t+2)=0` or

`t=1` or `t=-2`

Answer 2

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`.