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
5Â Â Â Â 10Â Â Â Â Â 30Â Â Â Â Â 150Â Â Â Â Â 950Â Â Â Â Â 11550
...7, 8, 12, 21, 37, ?
12 3 15 ? 153 756
...A series is 48, 129, 298, 587, 1028, 1653
If another series 150, ___, m, ___, follows the same pattern as the given number series, then find th...
1 9 64 255 755 1835
...100Â Â Â Â 221Â Â Â Â 77Â Â Â Â 245Â Â Â Â 50Â Â Â 275
8Â Â Â Â 20Â Â Â Â 36Â Â Â Â Â 56Â Â Â Â Â 80Â Â Â Â ?
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If   314     306      x      269       394    178    1480
Then, 1/3 x + 2x - x = ?
...6000 3002 1503 ? 378.75 191.375 97.6875
...