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4 years ago

13

Please help brainliest and 20 points!

1 answer:

Nadya [2.5K]4 years ago

4 0

Answer:

I'm so sorry I can't help haven't learned how to do that yet

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3 years ago

Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

LiRa [457]

we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

Let's verify each case to determine the solution to the problem

<u>case A)</u> $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> <u>the term is not a perfect cube</u>

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

therefore

The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

<u>case B)</u> $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> <u>the term is not a perfect cube</u>

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

<u>case C)</u> $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> <u>the term is not a perfect cube</u>

$a^{22}$ ------> <u>the term is not a perfect cube</u>

$b^{16}$ ------> <u>the term is not a perfect cube</u>

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

<u>case D)</u> $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> <u>the term is not a perfect cube</u>

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

<u>case E)</u> $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

5 0

3 years ago

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