What's the FASTEST Way to Master MATH in 2024? | Radical Equations

[@gregevgeni1864]

Another approach.
The given equation is equivalent to
³√(27-x²) +(-³√8) + (-³√(1-x²) = 0 or
A + B + C = 0 (1) ,
A=³√(27-x²), B=-³√8 , C=-³√(1-x²) (2)
From (*) => A³+B³+C³=3 ABC (3).
The (3) due to (2) rewrite as
(27-x²)+(-8)+(-(1-x²))=
3•³√((-8)(27-x²)(-(1-x²))) =>
6=³√(8(27-x²)(1-x²)) =>
216 = 8(27-28x²+x⁴) =>
x⁴-28x²=0 => x²(x²-28)=0 =>
x = 0 , x = ±2√7

[@gregevgeni1864]

x = ±2√7 , x = 0
The given equation is equivalent to
³√(27-x²) = 2+³√(1-x²) (1)
Let t=³√(27-x²) => t³=27-x² (2).
r=³√(1-x²) => r³=1-x² (3).
From (1), (2),(3) =>
t=2+r and t³-r³=26 .
Solving the system we have
r=-3 or r=1 .
From (3) and r=-3 => ³√(1-x²) =-3 =>
1-x²=-27 => x²=28 => x=±2√7.
From (3) and r=1 => ³√(1-x²)=1=>
1-x²=1 => x=0.

[@潘博宇-k4l]

X1=0, X2=+,-(28)^(1/2)=+,-2(7)^(1/2).

[@Fjfurufjdfjd]

χ=0 (διπλη) ή χ=+ 2(7)^(1/2) ή χ=-2(7)^(1/2).

[@ManojkantSamal]

^=reas as to the power
*=read as square root
As per question
{2+(5+x)(5-x)}^(1/3)=2+{(1+x)(1-x)}^(1/3)
So,
{2+25-x^2}^(1/3)=2+(1-x^2)^(1/3)
(27- x^2)^(1/3)=2+(1-x^2)^(1/3)
(27-x^2)^(1/3)-(1-x^2)^(1/3)=2
Let( 27-x^2)^(1/3)=R, (1-x^2)^(1/3)=L
So,
R-L=2
RL={(27-x^2)(1-x^2)}^(1/3)
={27-27x^2-x^2+x^4}^(1/3)
=(x^4-28x^2+27)^(1/3)
Now,
(R-L)^3=R^3-L^3-3RL(R-L)
2^3={27-x^2-1+x^2}-3RL(2)
8=26-6RL
-6RL=8-26=-18
RL=-18/-6=3
(RL)^3=3^3=27
X^4-28x^2+27=27
X^4-28x^2=27-27=0
X^4-28x^2=0
X^2(x^2-28)=0
So,
X^2=0 or x^2-28=0
X=0,or x^2=28
Now x^2=28
X=*(28)=2.*7
Hence
X=0 or 2.*7

[@RajeshKumar-wu7ox]

X=0,2√7 ,-2√7