How can I solve algebras
(2b3c/a3b2c2)4/6

 " " .

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Step-by-step explanation:

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We are asked to simplify the following given expression:

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             ;

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Let us start with the "numerator" ;  which functions as a separate "fraction"—in and of itself—with its own numerator and denominator:

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      ;

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Note the following "division" property of exponents:

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 →    ;   .

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Then:

→  (\frac{2b^3c}{a^3b^2c^2})^4 ;

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   =    ;   ; ; .

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Then, start by simplifying "this particular numerator":

     →   ;

Note the following "multiplication" property of exponents:

→  ;   .

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Then:

→ ;

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The continue by simplifying "this particular denominator":

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→

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And we have:

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→     .

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Note the following properties of exponents:

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→ ;   .

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→ ;   .

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So, we have:  

→ \frac{16b^1^2c^4}{a^1^2b^8c^8} ;

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→ 16 ÷ 1  = 16 ;  ("1" is the "implied coefficient") ; in the numerator.

→ ;  stays in the denominator;

→   ; replaces the original: " " [in the numerator];   is eliminated in the denominator;

→ ;  The original " " [in the numerator] is eliminated; the original " " [in the denominator] is replaced with " " . ________________

And the expression is rewritten as:

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→   ;

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→ Now, from the original given problem, we divide this value by "6" ;

which is the same value we get by multiplying this value by: " " ;

→ as follows:

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→    ;

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Considering the "16" and the "6" ;  each of these can be divided by "2" ;

Specifically, " (16 ÷2 = 8) " ;  and:  " (6 ÷ 2 = 3) " .

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So, we can rewrite the expression — by substituting "8" in lieu of the "16" ; and "3" in lieu of the "6" ; as follows:

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→    ;

And further simplify;

→

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Hope this helps!

  Best wishes!

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