Question

Difficulty: Very hardOperations with Variable Exponents

Simplify the following expression completely. Assume xx and yy are non-zero real numbers.

(3x2y−3)−2(−3x−2y2)3(x−4y3)−2 \frac{(3x^2y^{-3})^{-2} (-3x^{-2}y^2)^3}{(x^{-4}y^3)^{-2}}

If the simplified expression is written in the form AxByCAx^B y^C, where AA, BB, and CC are integers, what is the value of A+B+CA + B + C?

Answer: -3

Answer

-3
The correct answer is found by systematically applying the laws of exponents. First, distribute the outer exponents to all factors inside the parentheses, being careful with signs and coefficients. The numerator simplifies to (19x−4y6)⋅(−27x−6y6)=−3x−10y12(\frac{1}{9}x^{-4}y^6) \cdot (-27x^{-6}y^6) = -3x^{-10}y^{12}. The denominator simplifies to x8y−6x^8y^{-6}. Dividing the numerator by the denominator using the quotient rule yields −3x−18y18-3x^{-18}y^{18}. Setting this equal to the target form AxByCAx^By^C yields A=−3A = -3, B=−18B = -18, and C=18C = 18. The sum A+B+CA + B + C is −3−18+18=−3-3 - 18 + 18 = -3.

Step-by-Step Solution

1
Apply the power of a product rule (ab)n=anbn(ab)^n = a^n b^n and the power of a power rule (am)n=am⋅n(a^m)^n = a^{m \cdot n} to the first factor in the numerator.
(3x2y−3)−2=3−2⋅x2(−2)⋅y−3(−2)=19x−4y6(3x^2y^{-3})^{-2} = 3^{-2} \cdot x^{2(-2)} \cdot y^{-3(-2)} = \frac{1}{9}x^{-4}y^6
To eliminate the outer negative exponent on the first term.
2
Apply the same exponent rules to the second factor in the numerator.
(−3x−2y2)3=(−3)3⋅x−2(3)⋅y2(3)=−27x−6y6(-3x^{-2}y^2)^3 = (-3)^3 \cdot x^{-2(3)} \cdot y^{2(3)} = -27x^{-6}y^6
To eliminate the outer exponent on the second term.
3
Multiply the simplified factors in the numerator using the product rule am⋅an=am+na^m \cdot a^n = a^{m+n}.
(19x−4y6)⋅(−27x−6y6)=(19⋅−27)⋅x−4−6⋅y6+6=−3x−10y12(\frac{1}{9}x^{-4}y^6) \cdot (-27x^{-6}y^6) = (\frac{1}{9} \cdot -27) \cdot x^{-4-6} \cdot y^{6+6} = -3x^{-10}y^{12}
To combine the expressions in the numerator into a single term.
4
Apply the power rules to simplify the denominator.
(x−4y3)−2=x−4(−2)⋅y3(−2)=x8y−6(x^{-4}y^3)^{-2} = x^{-4(-2)} \cdot y^{3(-2)} = x^8y^{-6}
To eliminate the outer negative exponent in the denominator.
5
Divide the numerator by the denominator using the quotient rule aman=am−n\frac{a^m}{a^n} = a^{m-n}.
−3x−10y12x8y−6=−3⋅x−10−8⋅y12−(−6)=−3x−18y18\frac{-3x^{-10}y^{12}}{x^8y^{-6}} = -3 \cdot x^{-10-8} \cdot y^{12-(-6)} = -3x^{-18}y^{18}
To reach the fully simplified algebraic form.
6
Identify the values of AA, BB, and CC to calculate the required sum.
A=−3A = -3, B=−18B = -18, C=18C = 18. The sum is −3+(−18)+18=−3-3 + (-18) + 18 = -3.
To answer the specific question asked in the stem.

Key Concept

Applying multiple rules of exponents (power of a power, negative exponents, product and quotient rules) to simplify complex algebraic expressions.
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