Question

How do you factor \[216{x^3} + 1\]?

Answer 1

Hint: Here in this question, we have to find the factors of the given equation. If you see the equation it is in the form of \[{a^3} + {b^3}\]. We have a standard formula on this algebraic equation and it is given by \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], hence by substituting the value of a and b we find the factors

Complete step-by-step solution:
The equation is an algebraic equation or expression, where algebraic expression is a combination of variables and constant.
Now consider the given equation \[216{x^3} + 1\], let we write in the exponential form. The number \[216{x^3}\] can be written as \[6x \times 6x \times 6x\] and the \[1\]can be written as \[1 \times 1 \times 1\], in the exponential form it is \[{\left( 1 \right)^3}\]. The number \[216{x^3}\] is written as \[6x \times 6x \times 6x\] and in exponential form is \[{(6x)^3}\]. Therefore, the given equation is written as \[{\left( {6x} \right)^3} + {1^3}\], the equation is in the form of \[{a^3} + {b^3}\].The \[{a^3} + {b^3}\]have a standard formula on this algebraic equation and it is given by \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], here the value of a is \[6x\] and the value of b is 1. By substituting these values in the formula, we have
\[216{x^3} + 1 = {\left( {6x} \right)^3} + {1^3} = (6x + 1)({(6x)^2} - (6x)(1) + {1^2})\]
On simplifying we have
\[ \Rightarrow 216{x^3} + 1 = (6x + 1)(36{x^2} - 6x + 1)\]
The second term of the above equation can be solved further by using factorisation or by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Let we consider \[36{x^2} - 6x + 1\], and find factors for this. Here a=36, b=-6 and c=1. By substituting these values in the formula we get
\[x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(1)(36)} }}{{2(36)}}\]
On simplification we have
\[ \Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 144} }}{{72}}\]
\[ \Rightarrow x = \dfrac{{6 \pm \sqrt { - 108} }}{2}\]
On further simplifying we get an imaginary number so let us keep as it is.
Therefore, the factors of \[216{x^3} + 1\] is \[(6x + 1)(36{x^2} - 6x + 1)\]

Note: To find the factors for algebraic equations or expressions, it depends on the degree of the equation. If the equation contains a square then we have two factors. If the equation contains a cube then we have three factors. Here this equation also contains 3 factors, the two factors are imaginary.