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2 years ago

(5√3-√27)^3 how do you prove that this is an integer?

Mathematics

2 answers:

mr Goodwill [35]2 years ago

5 0

Keys:

$\left(a\cdot \:b\right)^n=a^nb^n$

Step-by-step explanation:

$\left(5\sqrt{3}-\sqrt{27}\right)^3\\\5\sqrt{3}-\sqrt{27}^3=\left(2\sqrt{3}\right)^3\\= 2\sqrt{3}\\\left(2\sqrt{3}\right)^3=2^3\left(\sqrt{3}\right)^3\\=2^3\left(\sqrt{3}\right)^3\\2^3=8\\=8\left(\sqrt{3}\right)^3\\=8\cdot \:3\sqrt{3}\\8\cdot \:3=24\\=24\sqrt{3}$

kirill [66]2 years ago

4 0

Answer:

It is not an integer.

Step-by-step explanation:

$(5\sqrt{3} -\sqrt{27} )^3\\\\=(\sqrt{3} *(5-3))^3\\\\=8*3*\sqrt{3} \\\\=24\sqrt{3} \\$

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Answer:

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