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a) Ta có hằng đẳng thức \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Vậy nên \(a^3+b^3+c^3+6=0.\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow a^3+b^3+c^3=-6.\)
b) \(x^3+y^3+3xy=x^3+3xy\left(x+y\right)+y^3=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=1.\)
c) \(x^3-y^3-3xy=x^3-3xy\left(x-y\right)-y^3=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=1.\)
a) (7x + 4)2 - (7x + 4)(7x - 4)
= 49x2 + 56x + 16 - 49x2 + 16
= 56x + 32
b) (x - 2y)3 - 6xy(x - 2y)
= x3 - 6x2y + 12xy2 - 8y3 - 6x2y + 12xy2
= x3 - 12x2y + 24xy2 - 8y3
c) (3x + y)(9x2 - 3xy + y2) - (3xy)3 - 27x2y
= 27x3 + y3 - (3xy)3 - 27x2y
d) 5(x + 3)(x - 3) + (2x + 3)2 + (x - 6)2
= 5x2 - 45 + 4x2 + 12x + 9 + x2 - 12x + 36
= 10x2
e) (2x + 3)2 + (2x - 3)2 - 2(4x2 - 9)
= (2x + 3)2 + (2x - 3)2 - 2(2x - 3)(2x + 3)
= (2x + 3 - 2x + 3)2
= 62 = 36
g) (x + 2)3 + (x - 2)3 + x3 - 3x(x - 2)(x + 2)
= (x+2+x-2)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4) + x3 - 3x3 + 12x
= 2x(x2 + 8) + x3 - 3x3 + 12x
= 2x3 + 16x + x3 - 3x3 + 12x
= 28x
a) \(x^2-3x+xy-3y\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
b) \(x^2+y^2-2xy-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y+5\right)\left(x+y-5\right)\)
c) \(4x^2-4xy+y^2=\left(2x-y\right)^2\)
m) \(81-x^2+2xy-y^2\)
\(=9^2-\left(x-y\right)^2\)
\(=\left(9-x+y\right)\left(9+x-y\right)\)
k) \(x^2-xy-x+y\)
\(=x\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
\(C=\left(x^3+y^3\right)+3xy\left(x^2+y^2+2xy\left(x+y\right)\right)\)
\(C=\left(x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2\right)+3xy\left(x^2+y^2+2xy\right)\) (vì x+y=1)
\(C=\left(x+y\right)^3-3x^2y-3xy^2+3xy\left(x+y\right)^2\)
\(C=1^3-3xy\left(x+y\right)+3xy.1^2\) (vì x+y=1)
\(C=1-3xy+3xy\)(vì x+y=1)
\(C=1\)
\(D=2\left(\left(x+y\right)^3-3xy\left(x+y\right)\right)-3\left(\left(x+y\right)^2-2xy\right)\)
\(D=2\left(1^3-3xy\right)-3\left(1^2-2xy\right)\)(vì x+y=1)
\(D=2-6xy-3+6xy\)
\(D=-1\)
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