Question

Factorise.
a⁶ + 5a² + 8

Do it quickly. Urgent.​

Answer 1

Over Reals

The range of the function is $a^6+5a^2+8\geq 8$, so the graph doesn't meet the x-axis above it.

Hence, no point of intersection or no real zero.

It is irreducible over reals.

Over Complex

Let's factorize $a^6+5a^2+8$ in a cubic polynomial.

Let $t=a^2$.

$\implies t^3+5t+8$

This is a depressed cubic, which quadratic coefficient is 0.

Let's attempt to find the solutions via Cardano's method.

Cardano's Method

Two main ideas are

• Every cubic can be expressed in a depressed cubic.
• $(a+b)^3-3ab(a+b)=a^3+b^3$

The second identity $(a+b)^3-3ab(a+b)=a^3+b^3$ is a well-known one.

It can explain that one of the solutions is $t=a+b$, then the equation we are solving is $t^3-3abt=a^3+b^3$.

Hence, solving the depressed cubic $t^3+px+q=0$ means finding $p$ and $q$ such that

• $p=3ab$
• $q=-a^3-b^3$

We deduce

• $a^3b^3=-\dfrac{p^3}{27}$
• $a^3+b^3=-q$

Now, we can find the quadratic equation, which roots are $a^3$ and $b^3$.

We can solve $X^2+qX-\dfrac{p^3}{27} =0$.

However, the solutions for $X$ are the perfect cube of $a$ and $b$.

(Let $a^3=\alpha$ and $b^3=\beta$ for convenience.)

• $a=\sqrt[3]{\alpha } \omega ,\sqrt[3]{\alpha } \omega ^2,\sqrt[3]{\alpha }$
• $b=\sqrt[3]{\beta } \omega ,\sqrt[3]{\beta } \omega ^2, \sqrt[3]{\beta }$

Since $p$ is a real number we find pairs of $a$ and $b$ via $p=3ab$, $\omega^3=1$.

• $a=\sqrt[3]{\alpha }$ and $b=\sqrt[3]{\beta }$
• $a=\sqrt[3]{\alpha } \omega$ and $b=\sqrt[3]{\beta } \omega^2$
• $a=\sqrt[3]{\alpha } \omega^2$ and $b=\sqrt[3]{\beta } \omega$

As $t=a+b$, our solutions are $t=\sqrt[3]{\alpha } +\sqrt[3]{\beta }$, $t=\sqrt[3]{\alpha }\omega +\sqrt[3]{\beta }\omega^2$, $t=\sqrt[3]{\alpha } \omega^2+\sqrt[3]{\beta } \omega$.

Application

Now we can use this to solve our problem.

Given: $t^3+5t+8$

General depressed cubic form: $t^3+pt+q$

• $p=5$
• $q=8$

Quadratic equation: $X^2+9X-\dfrac{125}{27} =0$

$\implies 27X^2+243X-125=0$

$\implies X =\dfrac{-243\pm \sqrt{72549} }{54}$

According to the method mentioned above, the solutions are $t=\sqrt[3]{\alpha } +\sqrt[3]{\beta }$, $t=\sqrt[3]{\alpha }\omega +\sqrt[3]{\beta }\omega^2$, or $t=\sqrt[3]{\alpha } \omega^2+\sqrt[3]{\beta } \omega$. / $\alpha$ and $\beta$ are two solutions for $X$.

Hence our solutions for $t$ are

• $t=\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}$ or
• $t=\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}\omega +\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}\omega^2$ or
• $t=\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}\omega ^2+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}\omega$.

Answer

After we substitute $t=a^2$ back in, the six factors are

$(a+\sqrt{\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}})$

$(a-\sqrt{\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}})$

$(a+\sqrt{\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}\omega+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}\omega^2})$

$(a-\sqrt{\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}\omega+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}\omega^2})$

$(a+\sqrt{\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}\omega^2+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}\omega})$

$(a-\sqrt{\sqrt[3]{\dfrac{-243+ \sqrt{72549} }{54}}\omega^2+\sqrt[3]{\dfrac{-243- \sqrt{72549} }{54}}\omega})$

Answer 2

Answer:

$\huge\sf\underline\red{Question}$

Factorise.

a⁶ + 5a² + 8

Do it quickly. Urgent.

$\huge\sf\underline\blue{Answer}$

$\tt \: {a}^{6} + 5 {a}^{2} + 8 \\ \tt { ({a}^{2}) }^{ 3} + 5 {a}^{3} + ( {2})^{3} \\ \tt { ({a}^{2}) }^{ 3} + ( {2})^{3} + 5 {a}^{3} \\ \tt { ({a}^{2}) }^{ 3} + ( {2})^{3} - {a}^{3} + 6 {a}^{3}$

$\tt { ({a}^{2}) }^{ 3} + ( {2})^{3} + ( - {a}^{3} ) - 3( {a}^{2} )(2)( - a) \\ \tt \: ({a}^{2} + 2 + ( - {a}^{2} ))(( {a}^{2} ) ^{2} + ( {2})^{2} + ( { - a})^{2} - ( {a}^{2} )(2) - 2( - a) - ( a)( {a}^{2} ))$

$\tt \: ({a}^{2} + 2 - a)( {a}^{4} + 4 + {a}^{2} - {2a}^{2} + 2a + {a}^{3} \\ \tt \: ({a}^{2} + a + 2)( {a}^{4} + {a}^{3} - {a}^{2} + 2a + 4)$

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