Question

a^3+b^3-c^3+3abc÷a+b-c
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Answer 1

Answer:

a³+b³-c³+3abc ÷ a+b-c

$= > \frac{a {}^{3} + b {}^{3} - c {}^{3} + 3 abc }{a + b - c} \\ = > \frac{a \times a \times a \times b \times b \times b \times ( - c) \times ( - c) \times ( - c) \times 3abc}{a + b - c} \\ = > {a}^{2} + {b}^{2} - {c}^{2} + 3abc$

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