`a) \root[3]{6+x}+\root[3]{10-x}=4`
Đặt `\root[3]{6+x}=a; \root[3]{10-x}=b`
`=> a+b=4` (1)
Lại có `a^3+b^3=6+x+10-x=16`(2)
Từ (1)(2) ta có hệ pt: `{(a+b=4),(a^3+b^3=16):}<=>{(a=4-b),(a^3+(4-a)^3=16(**)):}`
Giải pt `(**)`: `a^3+64-48a+12a^2-a^3=16`
`<=> 12a^2-48a+48=0`
`<=> a^2-4a+4=0`
`<=> (a-2)^2=0`
`<=> a=2 => b=4-a=2`
Khi đó `{(\root[3]{6+x}=2),(\root[3]{10-x}=2):}<=>{(6+x=8),(10-x=8):}<=>x=2`
Vậy `S={2}`
`b) x=\root[3]{x}`
`<=> x^3=x`
`<=> x^3-x=0`
`<=> x(x^2-1)=0`
`<=> x(x-1)(x+1)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy `S={0;+-1}`
`c) 2x^3=(x+1)^3`
`<=> \root[3]{2}x=x+1`
`<=> (\root[3]{2}-1)x=1`
`<=> x=1/(\root[3]{2}-1)`
`<=> x=\root[3]{4}+\root[3]{2}+1`
Vậy `S={\root[3]{4}+\root[3]{2}+1}`
`d) \root[3]{14+x}.\root[3]{14-x}=4`
`<=> (14+x)(14-x)=64`
`<=> 196-x^2=64`
`<=> x^2=132`
`<=> x=2\sqrt{33}`
Vậy `S={2\sqrt{33}}`
