Subtraction of Algebraic Expressions

$(3q+7p^2–2r^3+4)–(4p^2-2q+7r^3-3)=?$

The perimeter of triangle is $6p^2–4p+9$ and two of its sides are $p^2-2p+1$ and $3p^2-5p+3$. Find the third side of the triangle.

The two adjacent sides of a rectangle are $5x^2-3y^2$ and $x^2+2xy$. Find the perimeter.

What must be subtracted from $3a^2-6ab-3b^2-1$ to get $4a^2-7ab-4b^2+1$?

Subtract $4p^2+5q^2-6r^2+7$ from $3p^2-4q^2-5r^2-6$

Subtract $5y^4-3y^3+2y^2+y-1$ from $4y^4-2y^3-6y^2-y+5$

Subtract:

$x^3+3x^2–5x+4$ from $3x^3–x^2+2x–4$

Subtract: $–6p+q+3r+8$ from $p–2q–5r–8$

Subtract:

$2a–5b+2c–9$ from $3a–4b–c+6$

Subtract $–16p$ from $–11p$

Subtract $–2abc$ from $–8abc$

Subtract $–8pq$ from $6pq$

Subtract:

$2a^{2}b$ from $–5a^{2}b$