Verified Answer :

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:

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.

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