Simplify left parenthesis root of order three of c to the power of seven end power d to the power of four end power right parenthesis squared. c to the power of start fraction seven over three end fraction end power d to the power of start fraction four over three end fraction end power c to the power of start fraction nine over 49 end fraction end power d to the power of start fraction nine over 16 end fraction end power 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 c3d2

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}

i think its a and d

step-by-step explanation:

i saw this question before, dont know the asnwer srry

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