converting LHS so it would equal to RHS?

can someone help me in converting the first equation so it would equal to the second equation:

LHS = k^2 * (2k^2 - 1 ) + [ 2k + 1 ]^3

RHS = (k+1)^2 [ 2(k+1)^2 -1 ]

i have tried many different ways but had no success so please help me!!

thanks in advance! :)

Update:

THANKS FOR BOTH OF THE ANSWER!! :)

2 Answers

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  • 1 decade ago
    Best Answer

    realising that

    1 k^2 + 2k + 1 is (k+1)^2 keep a 2k +1 in the 2nd expression and expand it.

    2 Then make 2k^-1 common and add the rest

    to check later if we can simplify

    LHS

    = k^2(2k^2-1)+ (2k+1)(2k+1)^2 using 1

    = k^2 (2k^2-1) + (2k+1)(4k^2+4k+1)

    = k^(2k^2-1) + (2k+ 1)(2k^2- 1 + 2k^2 + 4k + 2) using 2

    = (2k^2-1)(k^2 + 2k + 1) + (2k+1)(2k^2+4k + 2)

    = (2k^2-1)(k+1)^2 + 2(2k+1)(k+1)^2

    = (k+1)^2(2k^2 - 1 + 4k+ 2)

    = (k+1)^2 (2k^2 + 4k + 2 - 1)

    = (k+1)^2(2(k^2+2k +1) - 1)

    = (k+1)^2[ 2(k+1)^2-1)

    = RHS

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