Question
If ab + bc + ca = 0 Find the value of
1/(a2 - bc) + 1/(b2 - ca) + 1/(c2 - ab)Solution
ab + bc + ca = 0 bc = - ab – ca ca = - ab – bc ab = -bc –ca Now, 1/(a2 - bc) + 1/(b2 - ca) + 1/(c2 - ab) = 1/(a2 - (-ab - ca)) + 1/(b2 - (-ab - bc)) + 1/(c2 - (-bc-ca)) =1/(a2 + ab+ca) + 1/(b2 + ab+bc) + 1/(c2 + bc+ca) = 1/(a(a + b + c)) + 1/(b(a + b + c)) + 1/(c(a + b + c)) = (bc + ca + ab)/(abc(a + b + c)) = 0/(abc(a + b + c)) = 0
More Algebra Questions
15.99% of 549.99 ÷ 11.17 = ? ÷ 20.15
74.91% of 639.95 – 599.98% of 45 + 119.987 = ?
(4.88 × 5.76)2 - ?2 = 39.89 × 19.86
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exactvalue.)
(1800.23 ÷ 29.98) + (816.32 ÷ 23.9) + 1634.11 = ?
1449.98 ÷ 50.48 × 10.12 = ? × 2.16
36.05 × 5.02 + 12.052 = ? + 9.09 × 4.04Â
(31.9)3 + (34.021)² - (16.11)3 - (42.98)² = ?
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