Simplifying Algebraic Expressions with Exponents
The given expression is:
\((8a^4x^2y^2)^2 \div a^6x^2y^{-1}\)
To solve it, we will first apply the power to the terms inside the parentheses and then divide by the terms outside the parentheses.
Applying the power of 2 to each term inside the parentheses:
\(= (8^2 \cdot (a^4)^2 \cdot (x^2)^2 \cdot (y^2)^2)\)
Calculating each term separately:
\(8^2 = 64\)
\((a^4)^2 = a^{4 \cdot 2} = a^8\)
\((x^2)^2 = x^{2 \cdot 2} = x^4\)
\((y^2)^2 = y^{2 \cdot 2} = y^4\)
So after applying the power we have:
\(= 64a^8x^4y^4\)
Now we will divide this by \(a^6x^2y^{-1}\):
\(= \frac{64a^8x^4y^4}{a^6x^2y^{-1}}\)
Dividing each term:
\(64\) is just a constant, so it stays as is.
For \(a\), we subtract exponents (when dividing with the same base, subtract the exponents): \(a^{8-6} = a^2\).
For \(x\), similarly: \(x^{4-2} = x^2\).
For \(y\), we add the exponents because one of them is negative: \(y^{4-(-1)} = y^{4+1} = y^5\).
So the simplified expression is:
\(= 64a^2x^2y^5\)
This is the final answer.
