How do you simplify ${(7 x^{2} y z)}^{3}$ $(7x^2yz)^3$ ?

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Answer 1 · 2017-07-18T08:46:09

$= 343 x^{6} y^{3} z^{3}$ $=343x^6y^3z^3$

The index shown applies to all the factors inside the bracket.

Recall the laws of indices:

${(x y z)}^{m} = x^{m} y^{m} z^{m} and {(x^{m})}^{n} = x^{m n}$ $(xyz)^m = x^my^mz^m" and "(x^m)^n = x^(mn)$

Using both of these laws gives:

${(7 x^{2} y z)}^{3} = 7^{3} x^{6} y^{3} z^{3}$ $(7x^2yz)^3 = 7^3x^6y^3z^3$

While $7^{3}$ $7^3$ is acceptable, you might want to find the actual value.

$7^{3} x^{6} y^{3} z^{3}$ $7^3x^6y^3z^3$

$= 343 x^{6} y^{3} z^{3}$ $=343x^6y^3z^3$

How do you simplify (7x2yz)3(7x^2yz)^3?