Question

(a+2b-3c)²+(a+2b-3c)²
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Answer 1

Answer:

2a²+4b²+6c²

Step-by-step explanation:

(a+2b-3c)² + (a+2b-3c)²

=2(a+2b-3c)²

=(2a+4b-6c)²

=2a²+4b²+6c²

Answer 2

Here's your answer...hope it helps you!

$(a+2b-3c)^{2} +(a+2b-3c)^{2}$

$=>[(a)^{2} +(2b)^{2} +(3c)^{2} -2(a)(2b)-2(2b)(3c)+2(a)(3c)]+[(a)^{2} +(2b)^{2} +(3c)^{2} -2(a)(2b)-2(2b)(3c)+2(a)(3c)]$

$=>(a^{2} +4b^{2} +9c^{2} -4ab-12bc+6ac)+(a^{2} +4b^{2} +9c^{2} -4ab-12bc+6ac)$

$=>a^{2} +4b^{2} +9c^{2} -4ab-12bc+6ac+ a^{2} +4b^{2} +9c^{2} -4ab-12bc+6ac$

$=>2a^{2} +8b^{2} +18c^{2} -8ab-24bc+12ac$

$=>2(a^{2} +4b^{2} +9c^{2} -4ab-12bc+6ac)$