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CBSE Class 9 Answered

Q-9 If a+c+c=5 and ab+bc+ca=15, find the value ofleft parenthesis a plus b right parenthesis to the power of 3 space end exponent plus left parenthesis b plus c right parenthesis to the power of 3 space end exponent plus left parenthesis c plus a right parenthesis to the power of 3 space end exponent minus 3 left parenthesis a plus b right parenthesis left parenthesis b plus c right parenthesis left parenthesis c plus a right parenthesis
Asked by kishor.kp14 | 18 Jun, 2019, 10:07: AM
answered-by-expert Expert Answer
(a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b +c)(c + a) = 2a3 + 2b3 +2c3 - 6abc
2a3 + 2b3 + 2c3 - 6abc = 2(a3 + b3 +c3 - 3abc) = 2(a + b+ c)[a2 + b2 +c2 - (ab +bc + ca)] -------- (I)
(a + b + c)2 = a2 +b2 +c2 + 2(ab +bc + ca)
52 = a2 +b2 + c2 + 30
25 - 30 = a2 + b2 + c2
- 5 = a2 +b2 + c2
Put the values in equation (I) we get


(a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b +c)(c + a) = 2(a + b+ c)[a2 + b2 +c2 - (ab +bc + ca)]
                                                                                = 2(5)(-5 - 15)
                                                                                =  2(5)(-20) = - 200
Answered by Sneha shidid | 18 Jun, 2019, 10:28: AM
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