\(a=0;1;2;3\) ở câu a

\(a=0;1;2;3;4;5;6;7\) ở câu b

\(a=0;1;2;3;4;5;6\) ở câu c

a) a = 3

b) b = 8

c) x = 1

d) ab = 23

a) \(N=8a^3-27b^3\)

\(=\left(2a\right)^3-\left(3b\right)^3\)

\(=\left(2a-3b\right)^3+18ab\left(2a-3b\right)\)

\(=5^3+18\cdot12\cdot5\)

\(=125+1080=1205\)

b) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)

\(=a^3+b^3+6a^2b^2+3a^3b+3ab^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+2ab+b^2\right)\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a+b\right)^2\)

\(=\left(a+b\right)^3+3ab\left(a+b\right)\left(a+b-1\right)\)

\(=1^3+3ab\cdot1\cdot0\)

\(=1\)

a ) \(N=8a^3-27b^3\)

\(\Leftrightarrow N=\left(2a-3b\right)\left(4x^2+6ab+9b^2\right)\)

\(\Leftrightarrow N=5\left(4x^2+9b^2+72\right)\)

Ta có : \(2a-3b=5\)

\(\Leftrightarrow4a^2+9b^2=25+6ab\)

Thay vào ta được : \(N=5\left(25+6ab+72\right)=845\)

b ) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)

\(\Leftrightarrow K=\left(a+b\right)^3-3ab\left(a+b\right)+6a^2b^2\left(a+b\right)+3ab\left(a+b\right)^2-6a^2b^2\)

\(\Leftrightarrow K=1-3ab+6a^2b^2+3ab-6a^2b^2=1\)

c ) \(P=\left(\dfrac{x}{4}\right)^3+\left(\dfrac{y}{2}\right)^3\)

\(\Leftrightarrow P=\left(\dfrac{x}{4}+\dfrac{y}{2}\right)^3-3\left[\left(\dfrac{x}{4}\right)^2\dfrac{y}{2}+\dfrac{x}{4}\left(\dfrac{y}{2}\right)^2\right]\)

\(\Leftrightarrow P=\left(\dfrac{2\left(x+2y\right)}{8}\right)^3-3\left[\dfrac{x^2y}{32}+\dfrac{xy^2}{16}\right]\)

\(\Leftrightarrow P=8-3xy\left(\dfrac{x+2y}{32}\right)\)

\(\Leftrightarrow P=8-3.4\left(\dfrac{8}{32}\right)=5\)

a,ta có : x^3+y^3-xy(x+y)=(x+y)(x^2+y^2-xy) -xy(x+y)=(x+y)(x^2+y^2-2xy=(x+y)(x-y)^2

(đpcm)

b)x^3-y^3+xy(x-y)=(x-y)(x^2+y^2+xy)+xy(x-y)=(x-y)(x^2+y^2+2xy)=(x-y)(x+y)^2 (đpcm)

c)(a+b)(a^2-ab+b^2)+(a-b)(a^2+ab+b^2)=(a^3+b^3)+(a^3-b^3)==2a^3 (đpcm)