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[tex]\text{$\dfrac{2\bigg(x^{7}-4x^{5}+8x^{4}+16x^{3}+32x^{2}+128\bigg)}{x^{8}+16x^{4}+256}$}[/tex]
[tex]\;[/tex]
[tex]\Large\textrm{\underline{\underline{Formula}}}[/tex]
Try to understand the following polynomial identities, as these are very critical in solving the problem.
1
The difference of perfect squares polynomial is given by
[tex]\boxed{\text{$(a+b)(a-b)=a^{2}-b^{2}$}}[/tex]
[tex]\;[/tex]
2
The sum of perfect cubes polynomial is given by
[tex]\boxed{\text{$(a+b)(a^{2}-ab+b^{2})=a^{3}+b^{3}$}}[/tex]
[tex]\;[/tex]
3
The difference of perfect cubes polynomial is given by
[tex]\boxed{\text{$(a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}$}}[/tex]
[tex]\;[/tex]
[tex]\Large\textrm{\underline{\underline{Solution}}}[/tex]
Firstly, [tex]\text{$x^{8}+16x^{4}+256$}[/tex] will be the common denominator.
[tex]\;[/tex]
The reason is
[tex]\text{$\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)$}[/tex]
[tex]\text{$=\bigg\{\bigg(x^2+4\bigg)^{2}-\bigg(2x\bigg)^{2}\bigg\}\bigg(x^{4}-4x^{2}+16\bigg)$}[/tex]
[tex]\text{$=\bigg\{\bigg(x^{4}+8x+16\bigg)-4x^{2}\bigg\}\bigg(x^{4}-4x^{2}+16\bigg)$}[/tex]
[tex]\text{$=\bigg(x^{4}+4x+16\bigg)\bigg(x^{4}-4x^{2}+16\bigg)$}[/tex]
[tex]\text{$=\bigg(x^{4}+16\bigg)^{2}-\bigg(4x^{2}\bigg)^{2}$}[/tex]
[tex]\text{$=\bigg(x^{8}+32x^{4}+256\bigg)-16x^{4}$}[/tex]
[tex]\text{$=x^{8}+16x^{4}+256$}[/tex]
[tex]\;[/tex]
Now, we are supposed to simplify the given fraction.
[tex]\;[/tex]
Consider
[tex]\text{$\dfrac{\bigg(x+2\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}{\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}$}[/tex]
[tex]\text{$=\dfrac{\bigg(x^{3}+8\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}{\bigg(x^{8}+16x^{4}+256\bigg)}$}[/tex]
[tex]\;[/tex]
Consider
[tex]\text{$\dfrac{\bigg(x-2\bigg)\bigg(x^{2}+2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}{\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}$}[/tex]
[tex]\text{$=\dfrac{\bigg(x^{3}-8\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}{\bigg(x^{8}+16x^{4}+256\bigg)}$}[/tex]
[tex]\;[/tex]
Consider
[tex]\text{$\dfrac{16\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)}{\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}$}[/tex]
[tex]\text{$=\dfrac{16\bigg\{\bigg(x^{2}+4\bigg)^{2}-\bigg(2x\bigg)^{2}\bigg\}}{\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}$}[/tex]
[tex]\text{$=\dfrac{16\bigg\{\bigg(x^{4}+8x^{2}+16\bigg)-4x^{2}\bigg\}}{\bigg(x^{2}+2x+4\bigg)\bigg(x^{2}-2x+4\bigg)\bigg(x^{4}-4x^{2}+16\bigg)}$}[/tex]
[tex]\text{$=\dfrac{16\bigg(x^{4}+4x^{2}+16\bigg)}{\bigg(x^{8}+16x^{4}+256\bigg)}$}[/tex]
[tex]\;[/tex]
Let's consider the sum of numerators.
[tex]\text{$\bigg(x^{3}+8\bigg)\bigg(x^{4}-4x^{2}+16\bigg)+\bigg(x^{3}-8\bigg)\bigg(x^{4}-4x^{2}+16\bigg)+16\bigg(x^{4}+4x^{2}+16\bigg)$}[/tex]
[tex]\text{$=x^{3}\bigg(2x^{4}-8x^{2}+32\bigg)+8\bigg(-8x^{2}\bigg)+16\bigg(x^{4}+4x^{2}+16\bigg)$}[/tex]
[tex]\text{$=\bigg(2x^{7}-8x^{5}+32x^{3}\bigg)+\bigg(-64x^{2}\bigg)+\bigg(16x^{4}+64x^{2}+256\bigg)$}[/tex]
[tex]\text{$=2x^{7}-8x^{5}+16x^{4}+32x^{3}+64x^{2}+256$}[/tex]
[tex]\;[/tex]
Hence-
[tex]\text{$\dfrac{2x^{7}-8x^{5}+16x^{4}+32x^{3}+64x^{2}+256}{x^{8}+16x^{4}+256}$}[/tex]
or -
[tex]\text{$\dfrac{2\bigg(x^{7}-4x^{5}+8x^{4}+16x^{3}+32x^{2}+128\bigg)}{x^{8}+16x^{4}+256}$}[/tex]
- is the wanted answer.