If $\alpha + \beta = - 2$ and ${\alpha ^3} + {\beta ^3} = - 56$ , then the quadratic equations whose roots are $\alpha $ and $\beta $ is
A) ${x^2} + 2x - 16 = 0$
B) ${x^2} + 2x + 15 = 0$
C) ${x^2} + 2x - 12 = 0$
D) ${x^2} + 2x - 8 = 0$

Hint: In this question, to find the quadratic equations, we have to find the product of roots. We will use the given values to find the product of roots. Then we will put the value of the sum of roots and the product of roots in the quadratic formula.
The formula of the quadratic equation is
${x^2} - (\alpha + \beta )x + \alpha \beta = 0$ .

Complete step by step solution:
Here we have been given
$\alpha + \beta = - 2$ and
${\alpha ^3} + {\beta ^3} = - 56$
We know the quadratic formula i.e.
${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$
By comparing here we have
$a = \alpha $
and
$b = \beta $
So by applying the formula we can write
${(\alpha + \beta )^3} = {\alpha ^3} + {\beta ^3} + 3\alpha \beta (\alpha + \beta )$
By putting the values from the given question we can write:
${( - 2)^3} = - 56 + 3\alpha \beta ( - 2)$
On multiplying the values we have:
$ - 8 = - 56 - 6\alpha \beta $
By arranging the similar terms together we can write:
$ - 6\alpha \beta = - 8 + 56 \Rightarrow - 6\alpha \beta = 48$
On further simplifying the values we have:
$\alpha \beta = \dfrac{{ - 48}}{6} = - 8$
It gives the product of roots i.e.
$\alpha \beta = - 8$ .
Now the quadratic formula is
${x^2} - (\alpha + \beta )x + \alpha \beta = 0$ .
So by putting the values in the formula we can write:
${x^2} - ( - 2)x - 8$
It gives,
${x^2} + 2x - 8 = 0$
Hence the correct option is (D) ${x^2} + 2x - 8 = 0$.

Note:
We should note that for quadratic equation
$a{x^2} + bx + c = 0$ , if $\alpha ,\beta $ are the roots then, we can write sum of the roots is
 $\alpha + \beta = \dfrac{{ - b}}{a}$
Similarly the product of the roots can be calculated as
$\alpha \beta = \dfrac{c}{a}$ .
We can always cross check or solve the problem with this formula.