Question

How do you simplify $\dfrac{{2{z^3} + 128}}{{16 + 8z + {z^2}}}$?

Answer 1

Hint: In this question we have to simplify the expression, we will do this by using the algebraic identities, for the numerator we will use the identity ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$, and for the denominator we will use the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, then further simplification we will get the required result.

Complete step by step solution:
Given function is $\dfrac{{2{z^3} + 128}}{{16 + 8z + {z^2}}}$,
Now first take 2 common in the numerator we get,
$ \Rightarrow \dfrac{{2\left( {{z^3} + 64} \right)}}{{16 + 8z + {z^2}}}$,
Now we simplify the expression we get,
$ \Rightarrow \dfrac{{2\left( {{{\left( z \right)}^3} + {{\left( 4 \right)}^3}} \right)}}{{{{\left( 4 \right)}^2} + 2\left( 4 \right)\left( z \right) + {{\left( z \right)}^2}}}$,
Now we can see that the numerator is in form ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$, and the denominator is in form ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$,
In the numerator, $a = z$ and $b = 4$ and in the denominator also $a = z$and $b = 4$, sol substituting the values in the expression we get,
$ \Rightarrow \dfrac{{2\left( {z + 4} \right)\left( {{z^2} - \left( 4 \right)\left( z \right) + {4^2}} \right)}}{{{{\left( {z + 4} \right)}^2}}}$,
Now simplifying we get,
$ \Rightarrow \dfrac{{2\left( {z + 4} \right)\left( {{z^2} - 4z + 16} \right)}}{{{{\left( {z + 4} \right)}^2}}}$,
Now eliminating the like terms we get,
$ \Rightarrow \dfrac{{2\left( {{z^2} - 4z + 16} \right)}}{{\left( {z + 4} \right)}}$,
So, the simplified form is $\dfrac{{2\left( {{z^2} - 4z + 16} \right)}}{{\left( {z + 4} \right)}}$,
Finally, state your restrictions on the variable. This can be done by setting the denominator expression to 0 and solving, then we get,
$ \Rightarrow {z^2} + 8z + 16 = 0$,
Now applying identity we get,
$ \Rightarrow {\left( {z + 4} \right)^2} = 0$,
Now taking out the square root we get,
$ \Rightarrow z + 4 = 0$,
Now subtract 4 from both of the equation we get,
$ \Rightarrow z + 4 - 4 = 0 - 4$,
Now simplifying we get,
$ \Rightarrow z = - 4$,
So, from the above we can say that $z \ne - 4$.
So, the simplified form is $\dfrac{{2\left( {{z^2} - 4z + 16} \right)}}{{\left( {z + 4} \right)}}$, where $z \ne - 4$.


$\therefore $The simplified form of the given expression $\dfrac{{2{z^3} + 128}}{{16 + 8z + {z^2}}}$will be equal to $\dfrac{{2\left( {{z^2} - 4z + 16} \right)}}{{\left( {z + 4} \right)}}$ where $z \ne - 4$.

Note:
Algebraic identities are algebraic equations which are always true for every value of variables in them. In an algebraic identity, the left-side of the equation is equal to the right-side of the equation, some of the algebraic identities that are commonly used are:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$,
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$,
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$,
${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$,
${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$,
${\left( {a + b} \right)^3} = {a^3} + 3ab\left( {a + b} \right) + {b^3}$,
${\left( {a - b} \right)^3} = {a^3} - 3ab\left( {a - b} \right) - {b^3}$.