Math Hacks to Master Radical Equations

[@davidseed2939]

interesting to note that x=0 can be seen feom inspection. this corresponds to. a=1, b=4. . and the other solution is a=4 , b=1 . I have seen this in several of these solutions. What is the defining factor?

[@RashmiRay-c1y]

This is tantamount to (x^3+1)^1/3 -5 = [(x-4){x^2+4x+16}]^1/3 = (x^3-64)^1/3. Let (x^3+1)^1/3 = and (x^3-64)^1/3 =b. Then, a-b=5 and a^3-b^3=65. If ab=t, we get 65=5(25+3t) and hence t=ab=-4. So, a=1,4 which yield x=0, 63^1/3.

[@Quest3669]

Eqn can written as (x^3+1)^1/3 - (x^3-4^3)^1/3= 5 Solving for real x= 0; (63)^1/3

[@Fjfurufjdfjd]

χ=0 ή χ=(63)^(1/3)

[@neerajgupta7856]

0 and cube root 63

[@gregevgeni1864]

Obviously x≠4. The given equation is equivalent to ³√[(x³+1)/(x-4)²]- -³√[((x-4)²(x+8)+48(x-4))/(x-4)²]=³√[125/(x-4)²] ³√(x³+1) - ³√((x-4)²(x+8)+48(x-4)) = 5 ³√(x³+1) - ³√(x³-64) =5 (1). Let a = ³√(x³+1) and b = ³√(x³-64) (*). From (*) and (1) => a - b = 5 and a³ - b³ = 65 => a=1 , a=4 . From (*) ³√(x³+1) = 1 => x=0 or ³√(x³+1)=4 => x=³√63.

[@walterwen2975]

Math Hacks to Master Radical Equations: ³√[(x³ + 1)/(x - 4)²] - ³√[125/(x - 4)²] = ³√[x + 8 + 48/(x - 4)], x ≠ 4, x ϵ R; x =? [³√(x³ + 1) - 5]/[³√(x - 4)²] = ³√[(x - 4)(x + 8) + 48]/³√(x - 4) = ³√[(x² + 4x + 16)/(x - 4)] ³√(x³ + 1) - 5 = ³√[(x - 4)(x² + 4x + 16)] = ³√(x³ - 16x + 16x - 64) = ³√(x³ - 64) [³√(x³ - 64)]³ = [³√(x³ + 1) - 5]³, x³ - 64 = (x³ + 1) - 5³ - 15[³√(x³ + 1)][³√(x³ + 1) - 5] {³√[(x³ + 1)(x³ - 64)]} = - 8, x⁶ - 63x³ - 64 = - 64, x³(x³ - 63) = 0 x³ = 0, x = 0 or x³ - 63 = 0, x³ = 63, x = ³√63 Answer check: ³√(x³ + 1) - 5 = ³√(x³ - 64) x = 0: 1 - 5 = - 4, ³√(- 64) = - 4; Confirmed x = ³√63: ³√(63 + 1) - 5 = 4 - 5 = - 1, ³√(63 - 64) = - 4; Confirmed Final answer: x = 0 or x = ³√63