Please solve 24-31

​Q24. Use (a –b)2 = a2 – 2ab + b2 to evaluate the following:

           (i) (99)                     (ii) (997)2                       (iii) (9.8)2.



Q25. By using suitable identities, evaluate the following:

         (i) (103)3                      (ii) (99)3                         (iii) (10.1)3


Q26. If 2a – b + c = 0 , prove that 4a2 – b2 + c2 + 4ac = 0.

              Hint. 2a – b + c = 0  2a + c = b  (2a + c)2 = b2.


Q27. If  a + b + 2c = 0. prove that a3  + b3 + 8c3 = 6abc.



Q28. If a + b + c = 0, then find the value of  a 2 b c + b 2 c a + c 2 a b .


Q29. If x + y = 4, then find the value of x3 + y3 + 12xy – 64.


Q30. Without actually calculating the cubes, find the values of :

               (i) (27)3 + (–17)3 + (–10)3                               (ii) (–28)3 + (15)3 + (13)3.


Q31. Using suitable identify, find the value of :

           86 × 86 × 86   + 14 × 14 × 14 86 × 86 - 86   × 14 + 14 × 14
 

Dear student
28. Given a+b+c=0a2bc+b2ca+c2ab=a3+b3+c3abcWe know that if a+b+c=0then a3+b3+c3=3abcSo, a3+b3+c3abc=3abcabc=3
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24) a. (100-1) ka whole square a=100 b=1 (a-b)whole sqaure = a ka square + 2ab + b ka square = 100 × 100 + 2 × 100 × 1 + 1 × 1 10000 + 200 + 1 = 10201 Like this only u can perform all the parts .
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In 25th question u can do like in a part (100 + 3)ka whole cube and know apply the identity of (a+b)ka whole cube .
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