Question

Choose one of the factors of 500x3 + 108y18
(500x to the 3rd + 108y to the 18th)

A. 25x2 + 15xy6 + 9y2
B. 5x + 3y6
C. All of the above
D. 6

Answer 1

Answer:

Option C

Therefore, one of the factors of $500x^3 +108y^\left (18\right )$ is $(5x+3y^6)$.

Step-by-step explanation:

Given: $500x^3 +108y^\left (18\right )$

the common factor from $500x^3$ and $108y^\left (18\right )$ is 4.

therefore, $4\cdot \left ( 125x^3+27y^\left (18\right ) \right )$

$4\cdot \left ( (5x)^3+(3y^6)^3 \right))$

Now, use the formula for above expression: $(a^3+b^3)=(a+b)(a^2-ab+b^2)$

here, $a=5x$ and $b=3y^6$

$( (5x)^3+(3y^6)^3 \right))$$=(5x+3y^6)(25x^2-15xy^6+9y^12)$

Therefore, we have

$500x^3 +108y^\left (18\right )$$=4\cdot (5x+3y^6)\left (25x^2-15xy^6+9y^\left ( 12 \right )  \right )$

Therefore, one of the factors of $500x^3 +108y^\left (18\right )$ is $(5x+3y^6)$.

Answer 2

Answer:

The correct option is B.

Step-by-step explanation:

The given expression is

$500x^3+108y^{18}$

Take out 4 as a common factor.

$500x^3+108y^{18}=4\times (125x^3+27(y^6)^3)$

$500x^3+108y^{18}=4\times (5^3x^3+3^3(y^6)^3)$

$500x^3+108y^{18}=4\times ((5x)^3+(3y^6)^3)$

Use the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$,

Here $a=5x$ and $b=3y^6$

$500x^3+108y^{18}=4\times (5x+3y^6)((5x)^2-(5x)(3y^6)+(3y^6)^2)$

$500x^3+108y^{18}=4\times (5x+3y^6)(25x^2-15xy^6+9y^{12})$

The factors of given expression are 4, $(5x+3y^6)$ and $(25x^2-15xy^6+9y^{12})$.

Therefore only option B is correct.