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Answer :
(i)Given: \((5x^2 – 6x) ÷ 3x\)
\(=\frac{5x^2 – 6x}{3x}=\frac{5x^2}{3x}-\frac{6x}{3x}\)
\(=\frac{5}2x-2=\frac{1}3(5x-6)\)
(ii)Given:\((3y^8 – 4y^6 + 5y^4) ÷ y^4\)
\(=\frac{3y^8 – 4y^6 + 5y^4}{y^4}=\frac{3y^8}{y^4}-\frac{4y^6 }{y^4}+\frac{5y^4 }{y^4}\)
\(=3y^4-4y^2+5\)
(iii)Given:\(8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x^2y^2z^2\)
\(=\frac{8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)}{4x^2y^2z^2}=\frac{8\times x^3y^2z^2 \times (x+y+z)}{4x^2y^2z^2}\)
\(=2(x+y+z)\)
(iv)Given: \((x^3 + 2x^2 + 3x) ÷ 2x\)
\(=\frac{x^3 + 2x^2 + 3x}{2x}=\frac{x \times (x^2+2x+3)}{2\times x}\)
\(=\frac{1}2(x^2+2x+3)\)
(v)Given:\((p^3q^6 – p^6q^3 – p^6q^3) ÷ p^3q^3\)
\(=\frac{p^3q^6 – p^6q^3 – p^6q^3}{p^3q^3}=\frac{p^3q^3(q^3-p^3)}{p^3q^3}\)
\(=q^3-p^3\)