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(x+1)(x+2)(x+3)(x+6)-3x2
(x+1)(x+1+1)(x+3)(x+3+3)-3x2
[(x+1)2 +(x+1)] [(x+3)2 +3(x+3)]-3x2
(x+1)2(x+3)2 +3(x+1)2(x+3) + (x+3)2(x+1) + 3(x+1)(x+3) - 3x2
(x+1)2(x+3)2 +2(x+1)(x+3)(2x+3) + 3(x+1)(x+3) - 3x2
(x+1)2(x+3)2 +2(x+1)(x+3)(2x+3) +(2x+3)2 -(2x+3)2 - 4x2
i didnt get the last and second last step plzz xplain
Dear ajinkya (x+1)(x+2)(x+3)(x+6)-3x2 (x+1)(x+1+1)(x+3)(x+3+3)-3x2 [(x+1)2 +(x+1)] [(x+3)2 +3(x+3)]-3x2 (x+1)2(x+3)2 +3(x+1)2(x+3) + (x+3)2(x+1) + 3(x+1)(x+3) - 3x2 here i have taken common (x+1)(x+3) from the second and third term (x+1)(x+3)[3(x+1) +(x+3)] = 2(x+1)(x+3)(2x+3) (x+1)2(x+3)2 +2(x+1)(x+3)(2x+3) + 3(x+1)(x+3) - 3x2
Dear ajinkya
here i have taken common (x+1)(x+3) from the second and third term (x+1)(x+3)[3(x+1) +(x+3)] = 2(x+1)(x+3)(2x+3)
now i have added and subtracted (2x+3)2 to make perfect square [(x+1)(x+3) +(2x+3)]2
(x+1)2(x+3)2 +2(x+1)(x+3)(2x+3) +(2x+3)2 -(2x+3)2 +3(x+1)(x+3)- 3x2 now simplify last 3 term it will become -4x2 [(x+1)(x+3) +(2x+3)]2 - 4x2 [(x+1)(x+3) +(2x+3) - 2x][(x+1)(x+3) +(2x+3) + 2x] [x2+4x+6][x2+8x+6] [x2+4x+6][x+4+√10][x+4-√10] Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
(x+1)2(x+3)2 +2(x+1)(x+3)(2x+3) +(2x+3)2 -(2x+3)2 +3(x+1)(x+3)- 3x2
[(x+1)(x+3) +(2x+3)]2 - 4x2
[(x+1)(x+3) +(2x+3) - 2x][(x+1)(x+3) +(2x+3) + 2x]
[x2+4x+6][x2+8x+6]
[x2+4x+6][x+4+√10][x+4-√10]
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
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