# How do you solve 27x^3 + 343 = 0?

Sep 18, 2016

$x = - \frac{7}{3}$

#### Explanation:

$27 {x}^{3} + 343 \text{ is a sum of cubes}$ and factorises, in general, as follows.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${\left(3 x\right)}^{3} = 27 {x}^{3} \text{ and " (7)^3=343rArra=3x" and } b = 7$

$\Rightarrow 27 {x}^{3} + 343 = \left(3 x + 7\right) \left(9 {x}^{2} - 21 x + 49\right)$

$\left(3 x + 7\right) \left(9 {x}^{2} - 21 x + 49\right) = 0$

solve : $3 x + 7 = 0 \Rightarrow x = - \frac{7}{3}$

solve : $9 {x}^{2} - 21 x + 49 = 0$

Before attempting to factorise check the $\textcolor{b l u e}{\text{discriminant}}$

here a = 9 , b = - 21 and c = 49

${b}^{2} - 4 a c = {\left(- 21\right)}^{2} - \left(4 \times 9 \times 49\right) = - 1323 < 0$

Hence this quadratic has no real roots.

The only real solution is therefore $x = - \frac{7}{3}$

Sep 18, 2016

$x = - \frac{7}{3}$

#### Explanation:

It is a real advantage to know all the powers up to 1000 by heart!

In such a case, we could go the conventional method of factoring and solving fo each factor equal to 0.

However, if you recognise that both 27 and 343 are cubes, as there is only one variable, we can also write as follows:

$27 {x}^{3} + 343 = 0$

$27 {x}^{3} = - 343$

${x}^{3} = - \frac{343}{27}$

$x = \sqrt[3]{- \frac{343}{27}}$

It is possible to have a cube root of a negative value.

$x = - \frac{7}{3}$