Answer:
c^{14/3}d^{8/3} OR c to the power of start fraction 14 over three end fraction end power d to the power of start fraction eight over three end fraction end power
Step-by-step explanation:
Given the expression (∛c)^7d⁴)², we are to simplify it. According to indices:
(ⁿ√a)^n= a^b/n and (a^m)^n = a^{mn}
Applying this in the question
(∛(c^7d⁴))²
=[∛(c^{7×2} × (d⁴)²]
=(∛(c^{14} × d^8)]
= [c^{14} × d^8]^1/3
= [c^{14/3} × d^{8×1/3}]
= c^{14/3} × d^{8/3}
= c^{14/3}d^{8/3}
Hence the resulting expression is c^{14/3}d^{8/3}
