Question

30 wa Bata do phir maaze lo
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Answer 1

[b(c-a)]^2 = 4[a(b-c)]c(a-b)]

(bc - ab)^2 = 4(ab - ac)(ac - bc)

(bc - ab)^2 = 4(ab - ac)(-bc + ac)

(bc - ab)^2 = 4(-acb^2 + bca^2 + abc^2 - a^2 c^2)

b^2 c^2 - 2acb^2 + a^2 b^2 = -4acb^2 + 4bca^2 + 4abc^2 - 4a^2 c^2

(bc)^2 + 2acb^2 + (2ac)^2 - 4bca^2 - 4abc^2 = 0

(bc)^2 + 2acb^2 - 4bca^2 - 4abc^2 + (2ac)^2 = 0

[(ab + bc) - 2ac][(ab + bc) - 2ac] = 0

ab + bc - 2ac = 0

2ac = ab + bc

Divide each term by abc

2/b = 1/c + 1/a

2/b = 1/a + 1/c

hope it helps uu...!!!

#Dramqueen⭐

Answer 2

in 4 the step ,, (a+b)2 identity is used