2 Answers

Prashant Sindhu · Answer 1 · 2019-06-23T07:35:13+0000

For this case, we must simplify by applying distributive property. By definition, the distributive property states that:

(a + b) (c + d) = ac + ad + bc + bd

So, if we have:

$(3m ^ 3- \frac {1} {2} y) ^ 2$ that can be written as:

$(3m ^ 3- \frac {1} {2} y) (3m ^ 3- \frac {1} {2} y)$

Considering that
$+ * - = -\ and\ - * - = +$

$(3m ^ 3) (3m ^ 3) - (3m ^ 3) (\frac {1} {2} y) - (\frac {1} {2} y) (3m ^ 3) + (\frac {1 } {2} y) (\frac {1} {2} y)$

To multiply powers of the same base, the same base is placed and the exponents are added:

$(3 ^ 2) (m) ^ 6 - (\frac {3} {2} m ^ 3y) - (\frac {3} {2} m ^ 3y) + ({\frac {1} {2}} ^ 2y ^ 2)$

Equal signs are added and the same sign is placed.

$9m ^ 6-2 (\frac {3} {2} m ^ 3y) + \frac {1} {4} y ^ 2$

So:

$(3m ^ 3- \frac {1} {2} y) (3m ^ 3- \frac {1} {2} y) = 9m ^ 6-2 (\frac {3} {2} m ^ 3y) + \frac {1} {4} y ^ 2$

Answer:

$(3m ^ 3- \frac {1} {2} y) (3m ^ 3- \frac {1} {2} y) = 9m ^ 6- 3m ^ 3y + \frac {1} {4} y ^ 2$

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