Question

(2a-3b+4c)^2+(2a+3b-4c)^2+2(2a-3b+4c)(2a+3b-4c)

Answer 1

this is (a+b)^2 form so the answer is (4a)^2= 16a^2

Answer 2

HERE IS YOUR ANSWER..⬇⬇

We know that,

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

By using this formula ,

We get,

$(2a - 3 b+ 4c) {}^{2} + (2a + 3b - 4c) {}^{2} + 2(2 a- 3b + 4c)(2 a+ 3 b- 4c) \\ \\ = (2a - 3 b+ 4c) {}^{2} + 2(2 a- 3b + 4c)(2 a+ 3 b- 4c) + (2a + 3b - 4c) {}^{2} \\ \\ = [(2a - 3 b+ 4c) + (2 a+ 3 b- 4c)] {}^{2} \\ \\ = (2a - 3 b+ 4c + 2 a+ 3 b- 4c) {}^{2} \\ \\ = (2 a+ 2a) {}^{2} \\ \\ = (4a) {}^{2} \\ \\ = 16a {}^{2}$

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