Question

How do you factor the expression $1-124{{x}^{3}}$?

Answer 1

Hint: The above given is not looking like a quadratic equation, but we will convert this given equation into quadratic equation or we will factor the above equation in quadratic form. A quadratic equation is any equation that can be written in the standard form as$a{{x}^{2}}+bx+c=0$ where x is an unknown and a, b, c are constants. Since, we cannot write the above given equation in standard form, so we will factor it and simplify it. Actually factorization consists of writing a number or another mathematical object as a product of several factors.

Complete step by step solution:
The given equation is:
$ 1-124{{x}^{3}}$
To solve the above question we will use the formula of difference of cubes identity which is ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{a}^{2}}{{x}^{2}} \right)$
Now here let`s suppose a is equal to 1 and b is equal to$\sqrt[3]{124}$, now we can rewrite the above given equation by using the formula, we get
$\Rightarrow {{1}^{3}}-{{\left( bx \right)}^{3}}=\left( 1-bx \right)\left( 1+bx+{{b}^{2}}{{x}^{2}} \right)$
Now putting the value of b in the above equation we get,
$\Rightarrow 1-124{{x}^{3}}=\left( 1-\sqrt[3]{124}x \right)\left( 1+\sqrt[3]{124}x+{{\left( \sqrt[3]{124} \right)}^{2}}{{x}^{2}} \right)$
Now we can see that the above equation cannot be factor into simpler linear factors. So by simply using the difference of cube identity we get the factor of the given equation.
Hence we get the factor of the above given equation $1-124{{x}^{3}}$is$\left( 1-\sqrt[3]{124}x \right)\left( 1+\sqrt[3]{124}x+{{\left( \sqrt[3]{124} \right)}^{2}}{{x}^{2}} \right)$.

Note: Here if we have 125 except 124 in the given question then we can write $125={{5}^{3}}$ then our factors of the given question become:
 $\Rightarrow 1-125{{x}^{3}}=\left( 1-5x \right)\left( 1+5x+25{{x}^{2}} \right)$
The given equation is not in a quadratic form but we get its factor in a quadratic factor$\left( 1-\sqrt[3]{124}x \right)\left( 1+\sqrt[3]{124}x+{{\left( \sqrt[3]{124} \right)}^{2}}{{x}^{2}} \right)$, in this the factor the$\left( 1+\sqrt[3]{124}x+{{\left( \sqrt[3]{124} \right)}^{2}}{{x}^{2}} \right)$is in a quadratic form.