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# Which of the following is the result when (6x^3y^3)^4 + (4x^4)^3y6 is

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Which of the following is the result when (6x^3y^3)^4 + (4x^4)^3y6 is [#permalink]

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10 Sep 2017, 05:03
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Difficulty:

35% (medium)

Question Stats:

84% (02:26) correct 16% (01:49) wrong based on 38 sessions

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Which of the following is the result when $$(6x^3y^3)^4 + (4x^4)^3y^6$$ is divided by $$(2x^4y^6)^2$$, if $$xy \neq 0$$?

A. $$2^23^4x^4 + 2^4x^4y^{(–6)}$$

B. $$3^2x^4 + 4^2x^4y^{(–6)}$$

C. $$2^{10}x^4y + 2^4x^4y^4$$

D. $$2^33^4x^4 + 4^3x^{12}y^{16}$$

E. $$3^2x^4 + 4^3x^{12}y^{16}$$

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Which of the following is the result when (6x^3y^3)^4 + (4x^4)^3y6 is [#permalink]

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10 Sep 2017, 10:00
Bunuel wrote:
Which of the following is the result when $$(6x^3y^3)^4 + (4x^4)^3y^6$$ is divided by $$(2x^4y^6)^2$$, if $$xy \neq 0$$?

A. $$2^23^4x^4 + 2^4x^4y^{(–6)}$$

B. $$3^2x^4 + 4^2x^4y^{(–6)}$$

C. $$2^{10}x^4y + 2^4x^4y^4$$

D. $$2^33^4x^4 + 4^3x^{12}y^{16}$$

E. $$3^2x^4 + 4^3x^{12}y^{16}$$

Writing this solution out makes it look a lot harder than it is.

Also, when solving, I combined steps that, for clarity's sake, I broke out here.

The concept is simply $$\frac{a^{x}}{a^{y}} = a^{(x-y)}$$

$$\frac{(6x^3y^3)^4 + (4x^4)^3y^6}{(2x^4y^6)^2$$ = ?

Split the numerator:

$$\frac{(6x^3y^3)^4}{(2x^4y^6)^2} + \frac{(4x^4)^3y^6}{(2x^4y^6)^2$$

Distribute the exponents

$$\frac{(6^4x^{12}y^{12})}{(2^2x^8y^{12})} + (\frac{(4^3x^{12}y^6)}{(2^2x^8y^{12})$$

In the numerators, factor $$6^4$$ and $$4^3$$ (they have factors of 2, and will be divisible by the 2 in the denominator) and then substitute the results into respective numerators

($$6^4$$) = $$2^43^4$$
($$4^3$$) = $$(2^2)^3 = 2^6$$

$$\frac{(2^43^4)x^{12}y^{12}}{2^2x^8y^{12}} + \frac{(2^6)x^{12}y^6}{2^2x^8y^{12}$$

Simplify - divide similar terms in the numerator and denominator of both fractions by subtracting exponents of identical bases

$$2^{(4-2)}3^4x^{(12-8)}y^{(12-12)} + 2^{(6-2)})x^{(12-8)}y^{(6-12)}$$ =

$$2^23^4x^4 + 2^4x^4y^{(–6)}$$

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Which of the following is the result when (6x^3y^3)^4 + (4x^4)^3y6 is [#permalink]

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20 Jun 2018, 06:13
Let x=y=1, then when $$(6x^3y^3)^4 + (4x^4)^3y^6$$ is divided by $$(2x^4y^6)^2$$ = $$(6^4+4^3)/2^2$$=$$2^2*3^4+2^4$$

Only A. $$2^23^4x^4 + 2^4x^4y^{(–6)}$$ gives us this result.

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Which of the following is the result when (6x^3y^3)^4 + (4x^4)^3y6 is   [#permalink] 20 Jun 2018, 06:13
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