`a) (x-3)^3 + (2x+1)^3 + (5-3x)^3 = 0`

`<=> (x-3)+(2x+1)+(5-3x)=0 `

`<=> 3(x-3)(2x+1)(5-3x) = 0 `

`<=> x-3=0` hoặc `2x+1=0` hoặc `5-3x=0`

`<=> x=3` hoặc  `x=-1/2` hoặc `x=5/3`

Vậy `x = {3; -1/2; 5/3}`

`b) (2x-1)^3 + (x-3)^3 + (5-3x)^3 = 8`

`<=> (2x-1)+(x-3)+(5-3x)=1`

`<=>1 -3(-7x^2 +20x -17) + 3(2x-1)(x-3)(5-3x) = 8`

`<=> 3(2x-1)(x-3)(5-3x) + 21x^2 -60x +52 = 8`

`<=> 3(2x-1)(x-3)(5-3x) + 21x^2 -60x +44 = 0`

`<=> 3(2x-1)(-3x^2+14x-15) + 21x^2 -60x +44 = 0`

`<=> 3(-6x^3 +28x^2 -31x +15) + 21x^2 -60x +44 = 0 `

`<=>`` -18x^3 +84x^2 -93x +45 +21x^2 -60x +44 = 0 `

`<=>`` -18x^3 +105x^2 -153x +89 = 0`

`<=>`` 18x^3 -105x^2 +153x -89 = 0`

`<=> x^3 - (105/18)x^2 + (153/18)x - 89/18 = 0`

`<=> x^3 - 35/6 x^2 + 17/2 x - 89/18 = 0`

`<=>x~~4,03`

Vậy `x~~4,03`

`c) (x-4)^3 + (2x+3)^3 = (3x-1)^3 `

`<=> (x-4)^3 + (2x+3)^3 - (3x-1)^3 = 0 `

`<=> (x-4)+(2x+3)-(3x-1)=0`

`<=> 3(x-4)(2x+3)(-(3x-1))=0`

`<=> x-4=0` hoặc `2x+3=0` hoặc `3x-1=0`

`<=> x=4` hoặc `x=-3/2` hoặc `x=1/3`

Vậy `x = {4;-3/2; 1/3}`

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Áp dụng BĐT phụ có :

`1)a^3+b^3+c^3 = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc`

`2)a^3 + b^3 - c^3 = 3abc`

`3)a^3+b^3+c^3=3abc`