Question

1/cho a+b+c=0 . cmr:  a^3+b^3+c^3=3abc

2/ Cho A=x⁶-6x⁵+6x⁴-6x³+6x²-6x+6.Tính giá trị của A khi x=5

Answer 1

1.ta có: 
x^3 + y^3 + z^3 - 3xyz = (x+y)^3 + z^3 - 3x^2y - 3xy^2 - 3xyz 
= (x+y)^3 + z^3 - 3xy(x + y + z) 
= (x+y+z)^3 - 3(x+y)^2.z - 3(x+y)z^2 - 3xy(x + y + z) 
= (x+y+z)^3 - 3(x+y)z(x+ y + z) - 3xy(x + y + z) 
=(x+y+z)[(x+y+z)^2 - 3(x+y)z - 3xy] 
với x+y+z = 0 => x^3 + y^3 + z^3 - 3xyz = 0 => x^3 + y^3 + z^3 = 3xyz

2.

x=5

=>6=x+1

=> A=x⁶-6x⁵+6x⁴-6x³+6x²-6x+6=x⁶-(x+1).x⁵+(x+1)x⁴-(x+1)x³+(x+1)x²-(x+1)x+(x+1)

=x⁶-x⁶-x⁵+x⁵-x⁴+x⁴-x³+x³-x²+x²-x+x+1

=1

vậy A=1 khi x=5

Answer 2

1,

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=3abc\)

2,

\(A=\left(x-1\right)\left(x-5\right)\left(x^4+x^2+1\right)+1\)

x=5 thì A=1

Answer 3

a+b+c=0 =>a+b= -c

=>  (a+b)^3=-c^3

=>a^3+b^3+c^3= -3ab(a+b)

=>a^3+b^3+c^3=3abc      (vì a+b=c)

=> dpcm