Question

full solution of this question.
fast​

Answer 1

Answer:

The answer is simple

Step-by-step explanation:

1). The given Question = (4a-b+2c)whole sqaure

we know the formula i.e., (a+b+c)whole square = ( a)2+(b)2+(c)2+2ab+2bc+2ca

let us assume that a= 4a,

b = -b,

c = 2c.

so substitute this values in the formula

Now,

=). (4a-b+2c)2

=). (4a)2+(-b)2+(2c)2+2*(4a)*(-b)+2*(-b)*(2c)+2*(2c)*(4a)

=). 8(a)2+(b)2+4(c)2-8ab-4bc+16ca (or) 8(a)2+(b)2+4(c)2-2(4ab-2bc-8ca).

2) the given question = (3a-5b-c)2

we know that (a+b+c)2 = (a)2+(b)2+(c)2+2ab+2bc+2ca

let us assume that a= 3a

b= -5b

c= -c

now..

according to formula...

=). (3a-5b-c)2

=). (3a)2+(-5b)2+(-c)2+2*(3a)*(-5b)+2*(-5b)*(-c)+2*(-c)*(3a)

=). 9(a)2+25(b)2+(c)2+2(-15ab)+2(5bc)-2(3ac)

=). 9(a)2+25(b)2+(c)2-30ab+10bc-6ac (or) 9(a)2+25(b)2+(c)2-2(15ab-5bc+3ac)..

Answer 2

Step-by-step explanation:

Hope this answer helps you .If you like this answer please give me a like.