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Black_prince [1.1K]
3 years ago
14

Choose one of the factors of 500x3 + 108y18

Mathematics
2 answers:
S_A_V [24]3 years ago
8 0

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.

sammy [17]3 years ago
7 0

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).








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