Question

(-a + 6b-20) with identity (a+b+c)=a²+b²+c²+2ab+2bc+2ca​

Answer 1

Step-by-step explanation:

( a² + b² + c² - ab - bc - ca )By multiplying it by 2 and dividing it… ... ( a² + b² - 2ab + b² + c² - 2bc + a² + c² - 2ca ) ÷ 2 = [ ( a - b )² ...

Missing: 6b- ‎| Must include: 6b-

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Answer 2

AnsweR :

$\sf \bold{ {a}^{2} + {36b}^{2} + 400 - 12ab - 240b + 40a}$

SolutioN :

$\boxed { \sf \pink{[Using \: Identity] :(a + b + c) = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}}$

$\sf( - a + 6b - 20) \\ \\ \sf \: a = - a \\ \sf \: b = 6a \\ \sf \: c = - 20$

$\sf[( - a)+ 6b + (- 20)] = {( - a)}^{2} + {(6b)}^{2} + {( - 20)}^{2} + 2( - a)(6b) + 2(6b)( - 20) + 2( - 20)( - a)$

$\sf {a}^{2} + {36b}^{2} + 400 - 12ab - 240b + 40a$