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kenny6666 [7]
3 years ago
5

How do you factor 2x^3+5y^3

Mathematics
1 answer:
Pavel [41]3 years ago
4 0
All you do is...
\mathrm{Apply\:sum\:of\:cubes\:rule:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)

2x^3+5y^3=\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right)
\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right) \ \textgreater \  Refine

\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(2^{\frac{2}{3}}x^2-\sqrt[3]{10}xy+5^{\frac{2}{3}}y^2\right)

Hope this helps!
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Simplify this: log√24+ log6​
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simplifying log\sqrt{24}+log6 we get log(12\sqrt{6})

Step-by-step explanation:

We need to simplify: log\sqrt{24}+log6

Solving:

Applying the rule:

log a + log b = log(ab)

log\sqrt{24}+log6\\=log(6\sqrt{24})\\\sqrt{24} = \sqrt{2*2*2*3} =\sqrt{2^2*6} =\sqrt{2^2} \sqrt{6}=2\sqrt{6}\\  Putting\,\,value:\\=log(6*2\sqrt{6})\\=log(12\sqrt{6})

So, simplifying log\sqrt{24}+log6 we get log(12\sqrt{6})

Keywords: Simplifying Logarithms

Learn more about Simplifying Logarithms at:

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  • brainly.com/question/5758530

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