ab+bc+ac=0 find the value of 1/a2 -bc+1/b2-ca+1/c2-ab(2 is sqaure)
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1/(a² - bc) + 1/(b² - ca) + 1/(c² - ab) = 0 if ab+bc+ac=0
Step-by-step explanation:
1/(a² - bc) + 1/(b² - ca) + 1/(c² - ab)
ab + bc + ca = 0
=> bc = -a(b + c)
=> -bc = a(b+c)
a² - bc = a² + a(b + c) = a(a + b + c)
Similarly
b² - ca = b(b + a + c) = b(a + b + c)
c² - ab = c(a + b + c)
= 1/a(a + b + c) + 1/b(a + b + c) + 1/c(a + b + c)
= bc/abc(a + b + c) + ac/abc(a + b + c) + ab/abc(a + b + c)
= (bc + ac + ab)/abc(a + b + c)
= (ab + bc + ac)/abc(a + b + c)
= 0/abc(a + b + c)
= 0
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