Question

(k3+3k2-k) (k3+3k2+k)

Answer 1

taking K Cube + 3 k square as a and k as b
we can use the identity a minus b into a + b equals to a square minus b square
K Cube + 3 k square cube equals to a plus b the whole cube A cube plus b cube plus 3 a b in bracket a + b therefore k raised to the power 9 + 27 K raised to the power 6 plus 9 keras to the power 5 in bracket kq + 3 k square- K square here is your answer hope this helps!!!

Answer 2

hey dear sweet friend here is your answer


=> k³(k³+3k²+k) 3k²(k³+3k²+k) -k(k³+3k²+k)

=> k^9+3k^6+k^4 × 3k^6+9k^4+3k^3 × -k^4-3k^3-k^2

=> k^9+3k^6+3k^6+k^4+9k^4-k^4 3k^3-3k^3-k²

=> k^9+6k^6+9k^4+(k²)


is your answer