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

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Answer 1 · 2016-06-09T12:26:20

#t=1# or #t=-2#

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 · 2016-06-09T12:33:29

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

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#.