Cho \({x^6} - 1 = \left( {x + A} \right)\left( {x + B} \right)\left( {{x^4} + {x^2} + C} \right)\), biết \(A,\,B,C\) là các số nguyên. Khi đó \(A + B + C\) bằng

\({\left( {x + 2y} \right)^3}\).

\({\left( {2x + y} \right)^3}\).

\({\left( {2x - y} \right)^3}\).

\({\left( {8x + y} \right)^3}\).

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = 5\left( {x - y} \right)\left( {x + y} \right)\).

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = \left( {5x - y} \right)\left( {x - 5y} \right)\).

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = \left( {x - y} \right)\left( {x + y} \right)\).

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = 5\left( {x - y} \right)\left( {x - 5y} \right)\).

\({x^2} - 6x + 9 = {\left( {x - 3} \right)^2}\).       

\(4{x^2} - 4xy + {y^2} = {\left( {2x - y} \right)^2}\).

\({x^2} + x + \dfrac{1}{4} = {\left( {x + \dfrac{1}{2}} \right)^2}\).        

\( - {x^2} - 2xy - {y^2} =  - {\left( {x - y} \right)^2}\).

\({\left( {xy + 2} \right)^3}\).

\({\left( {xy + 8} \right)^3}\).

\({x^3}{y^3} + 8\).

\({\left( {{x^3}{y^3} + 2} \right)^3}\).

\({\left( {a - 3} \right)^2}{\left( {a + 3} \right)^2}\).

\({\left( {a + 3} \right)^4}\).

\(\left( {{a^2} + 36a + 9} \right)\left( {{a^2} - 36a + 9} \right)\).

\({\left( {{a^2} + 9} \right)^2}\)

\(9{x^2} + 0,03x + 0,1\).

\(9{x^2} + 0,6x + 0,01\).

\(9{x^2} + 0,3x + 0,01\).

\(9{x^2} - 0,3x + 0,01\).

\(  \left( {\dfrac{{{x^2}}}{4} + 5y} \right)\left( {\dfrac{{{x^4}}}{4} - \dfrac{5}{4}{x^2}y + 5{y^2}} \right)\).

\(\left( {\dfrac{{{x^2}}}{4} - 5y} \right)\left( {\dfrac{{{x^4}}}{{16}} + \dfrac{5}{4}{x^2}y + 25{y^2}} \right)\).

\(  \left( {\dfrac{{{x^2}}}{4} + 5y} \right)\left( {\dfrac{{{x^4}}}{{16}} - \dfrac{5}{4}{x^2}y + 25{y^2}} \right)\).

\(\left( {\dfrac{{{x^2}}}{4} + 5y} \right)\left( {\dfrac{{{x^4}}}{{16}} - \dfrac{5}{2}{x^2}y + 25{y^2}} \right)\).