# 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! :)

THANKS FOR BOTH OF THE ANSWER!! :)

### 2 Answers

- Mein Hoon NaLv 71 decade agoBest 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