Question

(3r-2k)³+(3r+2k)³ simply it

Answer 1

Answer:

By the formula (a+b)³=a³+b³+3ab(a+b)

(a-b)³= a³- b³- 3ab(a - b)

(3r-2k)³=3r³-2k³-3(3r)(2k)(3r-2k)

(3r+2k)³=3r³+2k³+3(3r)(2k)(3r+2k)

=3r³-2k³-9r(2k)(3r-2k) +3r³+2k³+9r(2k)(3r+2k)

Here 3r³ ,-3r³,-2k³,2k³ are cancelled which gives zero

=0-0-9r(2k)(3r-2k) +9r(2k)(3r+2k)

=0-27r-4k+27r+4k=0

Here again -27r,-4k,27r,4k are cancelled

So the answer is zero

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