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Question
simplify. express your answer using positive exponents.\\(\frac{(3b^{5}c^{0}d)(6b^{0}c^{-1}d^{7})}{6b^{-1}c^{0}d^{0}}\\)
Step1: Simplify zero exponents
Any non-zero term to the power of 0 is 1: $c^0=1$, $b^0=1$, $d^0=1$
Substitute into the expression:
$\frac{(3b^5 \cdot 1 \cdot d)(6 \cdot 1 \cdot c^{-1}d^7)}{6b^{-1} \cdot 1 \cdot 1} = \frac{(3b^5d)(6c^{-1}d^7)}{6b^{-1}}$
Step2: Multiply numerator coefficients/terms
Multiply constants and add exponents of like bases:
Constants: $3 \times 6 = 18$
$b$-terms: $b^5$
$c$-terms: $c^{-1}$
$d$-terms: $d^{1+7}=d^8$
Numerator becomes: $18b^5c^{-1}d^8$
Expression now: $\frac{18b^5c^{-1}d^8}{6b^{-1}}$
Step3: Simplify coefficients and exponents
Divide constants: $\frac{18}{6}=3$
$b$-terms: $b^{5-(-1)}=b^{6}$ (subtract denominator exponent)
$c$-terms: $c^{-1}$ (no $c$ in denominator)
$d$-terms: $d^8$ (no $d$ in denominator)
Expression: $3b^6c^{-1}d^8$
Step4: Convert negative exponents to positive
$c^{-1}=\frac{1}{c}$, so rewrite:
$3b^6d^8 \cdot \frac{1}{c} = \frac{3b^6d^8}{c}$
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$\frac{3b^6d^8}{c}$