Mathematics: (4x ^3 + 7y ^3z ^4) ^2

I hope this helps you 1/21+1/28=1/t4/84+3/84=1/t5/84=1/tt=84/584/5 121 ??.84/5=1.21?=5/4...

(4x ^3 + 7y ^3z ^4) ^2 - id4373426991504, fashiondiva316

terriblexsiren

29.01.2022

16 $x^{6}$ + 56x³y³ $z^{4}$ + 49 $y^{6}$ $z^{8}$

Step-by-step explanation:

Note that we are using the rule of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$

Given

(4x³ + 7y³ $z^{4}$ )² = (4x³ + 7y³ $z^{4}$ )(4x³ + 7y³ $z^{4}$ )

Each term in the second factor is multiplied by each term in the first factor, that is

4x³ (4x³ + 7y³ $z^{4}$ ) + 7y³ $z^{4}$ ( 4x³ + 7y³ $z^{4}$ ) ← distribute both parenthesis

= 16 $x^{6}$ + 28x³y³ $z^{4}$ + 28x³y³ $z^{4}$ + 49 $y^{6}$ $z^{8}$ ← collect like terms

= 16 $x^{6}$ + 56x³y³ $z^{4}$ + 49 $y^{6}$ $z^{8}$

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