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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

Mathematics

2 answers:

LiRa [457]3 years ago

5 0

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

case A) $9w^{33} -y^{12}$

we know that

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

$w^{33}=(w^{11})^{3}$ ------> 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

case B) $18p^{15} -q^{21}$

we know that

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

$p^{15}=(p^{5})^{3}$ ------> 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

case C) $36a^{22} -b^{16}$

we know that

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

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

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

therefore

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

case D) $64c^{15} -a^{26}$

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

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

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

case E) $64c^{15} -d^{27}$

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

tigry1 [53]3 years ago

3 0

The expression $\boxed{64{c^{15}} - {d^{27}}}$ is a difference of cubes.

Further Explanation:

Given:

The options are as follows,

(a). $9{w^{33}} - {y^{12}}$

(b). $18{p^{15}} - {q^{21}}$

(c). $36{a^{22}} - {b^{16}}$

(d). $64{c^{15}} - {a^{26}}$

(e). $64{c^{15}} - {d^{27}}$

Calculation:

The cubic formula can be expressed as follows,

$\boxed{{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}$

The expression is $9{w^{33}} - {y^{12}}.$

9 is not a perfect cube of any number, ${w^{33}}$ can be written as ${\left( {{w^{11}}} \right)^3}$ and ${y^{12}}$ can be represents as ${\left( {{y^4}} \right)^3}.$

$9{w^{33}} - {y^{12}}$ cannot be written as the difference of cube. Option (a) is not correct.

The expression is $18{p^{15}} - {q^{21}}.$

18 is not a perfect cube of any number, ${p^{15}}$ can be written as ${\left( {{p^5}} \right)^3}$ and ${q^{21}}$ can be written as ${\left( {{q^7}} \right)^3}.$

$18{p^{15}} - {q^{21}}$ cannot be written as the difference of cube. Option (b) is not correct.

The expression is $36{a^{22}} - {b^{16}}.$

36 is not a perfect cube of any number, ${a^{22}}$ is not perfect cube and ${b^{16}}$ is not a perfect cube.

$36{a^{22}} - {b^{16}}$ cannot be written as the difference of cube. Option (c) is not correct.

The expression is $64{c^{15}} - {a^{26}}.$

64 can be written as ${\left( {{2^2}} \right)^3}, {a^{26}}$ is not perfect cube and ${c^{15}}$ can be written as ${\left( {{c^5}} \right)^3}.$

$64{c^{15}} - {a^{26}}$ cannot be written as the difference of cube. Option (d) is not correct.

The expression is $64{c^{15}} - {d^{27}}.$

64 can be written as ${\left( {{2^2}} \right)^3}, {d^{27}}$ can be written as ${\left( {{d^9}} \right)^3}$ and ${c^{15}}$ can be written as ${\left( {{c^5}} \right)^3}.$