Question

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(i) $(2a-5b-7c{)}^2$ (ii) $(-3a+4b-5c{)}^2$ (iii) $(\frac{1}{2}a-\frac{1}{4}b+2{)}^2$

Answer 1

1-
$(2a-5b-7c{)}^2=(2a+(-5b)+(-7c){)}^2=(2a{)}^2+(-5b{)}^2+(-7c{)}^2+2(2a\left)\right(-5b)+2(-5b\left)\right(-7c)+2(2a\left)\right(-7c)=4a^2+25b^2+49c^2-20ab+70bc-28ac$

2-

=> (-3a+4b-5c)²

It is in the form of-----

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca


a=-3a , b=4b , c=-5c

= (-3a)²+(4b)²+(-5c)²+2(-3a)(4b) +2(4b)(-5c)+2(-5c)(-3a)

= 9a²+16b²+25c²-24ab-40bc+30ca

3 )

$(\frac{1}{2}a-\frac{1}{4}b+2{)}^2=(\frac{1}{2}a{)}^2+(-\frac{1}{4}b{)}^2+2^2+2(\frac{1}{2}a\left)\right(-\frac{1}{4}b)+2(-\frac{1}{4}b)2+2(\frac{1}{2}a)2=(\frac{a^2}{4})+\frac{b^2}{16}+4-\frac{ab}{4}-b+2a$