Yesterday i was doing the question from class 9th ncert book of math of Polynomials chapter of exercise 2.5. What is the easiest way for solving it because i was not able to do this question please help me for solving this question Without actually calculating the cubes, find the value of each of the following:(ii) (28)3+(−15)3+(−13)3
AnilSinghBoraGuru
Without actually calculating the cubes, find the value of each of the following:(ii) (28)3+(−15)3+(−13)3 Q.14(2)
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(28)3+(−15)3+(−13)3
Let a = 28
b = −15
c = −13
We know that if x+y+z = 0, then x3+y3+z3 = 3xyz.
Here, x+y+z = 28–15–13 = 0
(28)3+(−15)3+(−13)3 = 3xyz
= 0+3(28)(−15)(−13)
= 16380