Question

(7a+3b+4c)²
$(7a + 3b + 4c) {?}^{2}$
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Answer 1

Answer:

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$( {a + b + c})^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

so, (7a)^2+(3b)^2+(4c)^2+2(7a*3b+3b*4c+4c*7a)

$49 {a}^{2} + 9 {b}^{2} + 16 {c}^{2} + 2(21ab + 12bc + 28ac)$