What must be subtracted from 3a2-6ab-3b2-1 to get 4a2 -7ab-4b2+1?pls w

1.	What must be subtracted from 3a2-6ab-3b2-1 to get 4a2 -7ab-4b2+1?pls write in steps the numbers next to the alphabets are the squares and cube
Answer» $\huge\bold{\gray{\sf{Answer:}}}$ Explanation: Let the EQUATION be 'X' which be SUBTRACTED from $\bold{3 {a}^{2} - 6ab - 3 {b}^{2} - 1}$ To get $\bold{\red{4 {a}^{2} - 7ab - 4 {b}^{2} + 1}}$ ACCORDING to the QUESTION: $➠ 3 {a}^{2} - 6ab - 3 {b}^{2} - 1 - x = 4 {a}^{2} - 7ab - 4 {b}^{2} + 1$ $➠ - x = 4 {a}^{2} - 7ab - 4 {b}^{2} + 1 - (3 {a}^{2} - 6ab - 3 {b}^{2} - 1)$ $➠ - x = 4 {a}^{2} - 7ab - 4 {b}^{2} + 1 - 3 {a}^{2} + 6ab + 3 {b}^{2} + 1$ $➠ - x = 4 {a}^{2} - 3 {a}^{2} - 7ab + 6ab - 4 {b}^{2} + 3 {b}^{2} + 1 + 1$ $➠ - x = {a}^{2} - ab - {b}^{2} + 2$ $➠ x = - ( {a}^{2} - ab - {b}^{2} + 2)$ $\bold{\boxed{\bold{\pink{x = - {a}^{2} + ab + {b}^{2} - 2}}}}$ $\huge\mathbb{CHECK}$ $3 {a}^{2} - 6ab - 3 {b}^{2} - 1 - ( - {a}^{2} + ab + {b}^{2} - 2)$ $➠ 3 {a}^{2} - 6ab - 3 {b}^{2} - 1 + {a}^{2} - ab - {b}^{2} + 2$ $\bold{\red{ ➠ 4 {a}^{2} - 7ab - 4 {b}^{2} + 1}}$