How do you solve # 5^3=(x+2)^3#?

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Answer 1 · 2015-09-10T02:07:28

#x = 3#

#5^3 = (x + 2)^3# => rewrite as:

#(x + 2)^3 - 5^3 = 0# => factor by the difference of cubes formula:

#[(x + 2) - 5][(x + 2)^2+ 5(x + 2)+25]=0# => simplify:

#(x - 3)(x^2 + 9x + 39)=0# => equate each bracket to zero:

#x - 3 = 0 => x = 3#

#x^2 + 9x + 39 = 0#

Discriminant #= b^2 - 4ac = 81 - 156 = -75<0#

Since the discriminant is negative the quadratic has no real solutions therefore the only valid solution is:

#x = 3#