Question

(3r-2k)cube + (3r+2k)cube how to solve

Answer 1

according to formulae we have:

${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \\ \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)$
now,


${(3r - 2k)}^{3} = 27 {r}^{3} - 8 {k}^{3} - 3rk(3r - 2k) \\ = > 27 {r}^{3} - 8 {k}^{3} - 9 {r}^{2} k + 6r {k}^{2}$
and

${(3r + 2k)}^{3} = 27 {r}^{3} + 8 {k}^{3} + 9 {r}^{2} k + 6r {k}^{2}$
now,

$27 {r}^{3} + 8 {k}^{3} + 9 {r}^{2} k + 6r {k}^{2} + 27 {r}^{3} - 8 {k}^{3} - 9 {r}^{2} k - 6r {k}^{2} \\ \\ = > 54 {r}^{3}$

Answer 2

Answer:

(3r-2k)3-(3r+2k)3

Step-by-step explanation: