Answer:
[tex]\mathrm{Factor}\:\frac{1}{27}x^{12}+\frac{8}{125}y^{18}:\quad \frac{x^{12}}{27}+\frac{8y^{18}}{125}[/tex]
Step-by-step Explanation:
Considering the expression
[tex]\frac{1}{27}x^{12}+\frac{8}{125}y^{18}[/tex]
[tex]\mathrm{Factor}\:\frac{1}{27}x^{12}+\frac{8}{125}y^{18}:[/tex]
First solving
[tex]\frac{1}{27}x^{12}[/tex]
[tex]\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}[/tex]
[tex]=\frac{1\cdot \:x^{12}}{27}[/tex]
[tex]\mathrm{Multiply:}\:1\cdot \:x^{12}=x^{12}[/tex]
[tex]=\frac{x^{12}}{27}[/tex]
Also solving
[tex]\frac{8}{125}y^{18}[/tex]
[tex]\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}[/tex]
[tex]=\frac{8y^{18}}{125}[/tex]
so equation becomes
[tex]\frac{1}{27}x^{12}+\frac{8}{125}y^{18}=\q\frac{x^{12}}{27}+\frac{8y^{18}}{125}[/tex]
Therefore,
[tex]\mathrm{Factor}\:\frac{1}{27}x^{12}+\frac{8}{125}y^{18}:\quad \frac{x^{12}}{27}+\frac{8y^{18}}{125}[/tex]
Answer:
[tex]\mathrm{Factor}\:\:\:\:\:27-\frac{1}{8}y^9:\quad \:-\left(\frac{y^3}{2}-3\right)\left(\frac{y^6}{4}+\frac{3y^3}{2}+9\right)[/tex]
Step-by-step Explanation:
Considering the expression
[tex]\mathrm{Factor\:out\:common\:term\:}-1[/tex]
[tex]=-\left(\frac{y^9}{8}-27\right)[/tex]
[tex]\mathrm{Factor}\:\frac{y^9}{8}-27:\quad \left(\frac{y^3}{2}-3\right)\left(\left(\frac{1}{2}\right)^2y^6+3\cdot \frac{1}{2}y^3+3^2\right)[/tex]
so
[tex]=-\left(\frac{y^3}{2}-3\right)\left(\left(\frac{1}{2}\right)^2y^6+3\cdot \frac{1}{2}y^3+3^2\right)[/tex]
[tex]\mathrm{Refine}[/tex]
[tex]=-\left(\frac{y^3}{2}-3\right)\left(\frac{y^6}{4}+\frac{3y^3}{2}+9\right)[/tex]
Therefore,
[tex]\mathrm{Factor}\:\:\:\:\:27-\frac{1}{8}y^9:\quad \:-\left(\frac{y^3}{2}-3\right)\left(\frac{y^6}{4}+\frac{3y^3}{2}+9\right)[/tex]
Answer:
[tex]\mathrm{Factor}\:\:\:\:\:c^{30}+\frac{27}{1000}:\:\:\left(c^{10}+\frac{3}{10}\right)\left(c^{20}-\frac{3c^{10}}{10}+\frac{9}{100}\right)[/tex]
Step-by-step Explanation:
Considering the expression
[tex]c^{30}+\frac{27}{1000}[/tex]
[tex]\mathrm{Factor}\:c^{30}+\frac{27}{1000}:\quad[/tex]
[tex]\mathrm{Rewrite\:}c^{30}+\frac{27}{1000}\mathrm{\:as\:}\left(c^{10}\right)^3+\left(\sqrt[3]{\frac{27}{1000}}\right)^3[/tex]
[tex]=\left(c^{10}\right)^3+\left(\sqrt[3]{\frac{27}{1000}}\right)^3[/tex]
[tex]\mathrm{Apply\:Sum\:of\:Cubes\:Formula:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)[/tex]
[tex]\left(c^{10}\right)^3+\left(\sqrt[3]{\frac{27}{1000}}\right)^3=\left(c^{10}+\frac{3}{10}\right)\left(c^{20}-\frac{3c^{10}}{10}+\left(\frac{3}{10}\right)^2\right)[/tex]
[tex]=\left(c^{10}+\frac{3}{10}\right)\left(c^{20}-\frac{3c^{10}}{10}+\left(\frac{3}{10}\right)^2\right)[/tex]
[tex]\mathrm{Refine}[/tex]
[tex]=\left(c^{10}+\frac{3}{10}\right)\left(c^{20}-\frac{3c^{10}}{10}+\frac{9}{100}\right)[/tex]
Therefore,
[tex]\mathrm{Factor}\:\:\:\:\:c^{30}+\frac{27}{1000}:\:\:\left(c^{10}+\frac{3}{10}\right)\left(c^{20}-\frac{3c^{10}}{10}+\frac{9}{100}\right)[/tex]
