Đáp án:

$\begin{array}{l}
1){\left( {7x + 4} \right)^2} - \left( {7x + 4} \right)\left( {7x - 4} \right)\\
 = \left( {7x + 4} \right)\left( {7x + 4 - 7x + 4} \right)\\
 = \left( {7x + 4} \right).8\\
 = 56x + 32\\
2){\left( {x + 2y} \right)^3} - 6xy\left( {x + 2y} \right)\\
 = \left( {x + 2y} \right)\left( {{{\left( {x + 2y} \right)}^2} - 6xy} \right)\\
 = \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\
 = {x^3} + 8{y^3}\\
3)\left( {3x + y} \right)\left( {9{x^2} - 3xy + {y^2}} \right)\\
 - {\left( {3x - y} \right)^3} - 27{x^2}y\\
 = 27{x^3} + {y^3} - \left( {27{x^3} - 27{x^2}y + 9x{y^2} - {y^3}} \right) - 27{x^2}y\\
 =  - 9x{y^2}\\
4)x\left( {x + 4} \right)\left( {x - 4} \right) - \left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)\\
 = x\left( {{x^2} - 16} \right) - {x^4} + 1\\
 = {x^3} - 16x - {x^4} + 1\\
5)\left( {y - 3} \right)\left( {y + 3} \right)\left( {{y^2} + 9} \right) - \left( {{y^2} + 2} \right)\left( {{y^2} - 2} \right)\\
 = \left( {{y^2} - 9} \right)\left( {{y^2} + 9} \right) - \left( {{y^4} - 4} \right)\\
 = {y^4} - 81 - {y^4} + 4\\
 =  - 77\\
6){\left( {a + b + c} \right)^2} - {\left( {a - c} \right)^2} - 2ab + 2bc\\
 = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac\\
 - {a^2} + 2ac - {c^2} - 2ab + 2bc\\
 = {b^2} + 4ac + 4bc\\
7){\left( {a + b + c} \right)^2} + {\left( {b + c - a} \right)^2}\\
 + {\left( {c + a - b} \right)^2} + {\left( {a + b - c} \right)^2}\\
 = {a^2} + 2a\left( {b + c} \right) + {\left( {b + c} \right)^2} + {\left( {b + c} \right)^2} - 2a\left( {b + c} \right) + {a^2}\\
 + {a^2} - 2a\left( {b - c} \right) + {\left( {b - c} \right)^2} + {a^2} + 2a\left( {b - c} \right) + {\left( {b - c} \right)^2}\\
 = 4{a^2} + 2{\left( {b + c} \right)^2} + 2{\left( {b - c} \right)^2}\\
 = 4{a^2} + 2{b^2} + 4bc + 2{c^2} + 2{b^2} - 4bc + 2{c^2}\\
 = 4{a^2} + 4{b^2} + 4{c^2}\\
8){\left( {a + b} \right)^3} + {\left( {a - b} \right)^3} - 6{a^2}b\\
 = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\\
 + {a^3} - 3{a^2}b + 3a{b^2} - {b^3} - 6{a^2}b\\
 = 2{a^3} + 6a{b^2} - 6{a^2}b\\
9){\left( {{x^2} - 1} \right)^3} - \left( {{x^4} + {x^2} + 1} \right)\left( {{x^2} - 1} \right)\\
 = {x^6} - 3{x^4} + 3{x^2} - 1 - {x^6} + 1\\
 =  - 3{x^4} + 3{x^2}\\
10)\\
{\left( {a + b} \right)^3} - {\left( {a - b} \right)^3} - 6{a^2}b\\
 = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\\
 - \left( {{a^3} - 3{a^2}b + 3a{b^2} - {b^3}} \right) - 6{a^2}b\\
 = 2{b^3}
\end{array}$