How do you solve $5^{3} = {(x + 2)}^{3}$ $5^3=(x+2)^3$ ?

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Answer 1 · 2015-09-10T02:07:28

$x = 3$ $x = 3$

$5^{3} = {(x + 2)}^{3}$ $5^3 = (x + 2)^3$ => rewrite as:

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

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

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

$x - 3 = 0 \Rightarrow x = 3$ $x - 3 = 0 => x = 3$

$x^{2} + 9 x + 39 = 0$ $x^2 + 9x + 39 = 0$

Discriminant $= b^{2} - 4 a c = 81 - 156 = - 75 < 0$ $= b^2 - 4ac = 81 - 156 = -75<0$

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

$x = 3$ $x = 3$

How do you solve 53=(x+2)3 5^3=(x+2)^3?