Harvard University Entrance top tricks || Admission Algebra Aptitude Test || Find x=?

[@mikeeisler6463]

Two variable substitution is not needed. Instead: a^3 = 4x + 76 ==> a^3 - 152 = 4x - 76 ==> a - (a^3 - 152)^(1/3) = 2 ==> (a - 2)^3 = a^3 - 152 ==> -6a^2 + 12a - 8 = -152 ==> a^2 - 2a - 24 = 0 a = 6 or a = -4 x = 35 or x = -35

[@RyanLewis-Johnson-wq6xs]

Surd[(4x+76),3]-Surd[(4x-76),3]=2 x= ± 35

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@thichhochoi766]

No need to solve for n. Less prone for error.

[@walterwen2975]

Harvard University Entrance top tricks: ³√(4x + 76) - ³√(4x - 76) = 2; x =? ³√(4x + 76) - ³√(4x - 76) = ³√[4(x + 19)] - ³√[4(x - 19)] = 2 ³√(x + 19) - ³√(x - 19) = 2/(³√4) = ³√(8/4) = ³√2 Let: u = ³√(x + 19), v = ³√(x - 19); u - v = ³√2, u³ - v³ = 38 u³ - v³ = (u - v)(u² + uv + v²) = (u - v)[(u - v)² + 3uv] 38 = (³√2)[(³√2)² + 3uv] = 2 + 3(³√2)uv, 3(³√2)uv = 38 - 2 = 36, uv = 12/(³√2) = 6(³√4) [³√(x + 19)][³√(x - 19)] = ³√(x² - 19²) = 6(³√4), x² - 19² = 4(6³), x² = 1225; x = ± 35 Answer check: ³√(4x + 76) - ³√(4x - 76) = ³√[4(x + 19)] - ³√[4(x - 19)] x = 35: ³√[4(54)] - ³√[4(16)] = ³√(6³) - ³√(4³) = 6 - 4 = 2; Confirmed x = - 35: ³√[4(- 16)] - ³√[4(- 54)] = ³√(- 4³) - ³√(- 6³) = - 4 + 6 = 2; Confirmed Final answer: x = 35 or x = - 35

[@shrikrishnagokhale3557]

My answer is x=+ or - 35

[@superacademy247]

That's it!💡💕😎

[@RealQinnMalloryu4]

{x^3+x^3 ➖ } {4x+4x ➖ }={x^6+8x^2}+76=8x^8 +{76+76 ➖}=152{ 8x^8+152}=160x^8 4^40x^8 4^4^10x^8 2^2^2^2^2^5x^2^3 1^1^1^1^1^1x1^1^2 x^1^2 (x ➖ 2x+1) (x^3 )^2➖( 4x)^2={x^9 ➖ 16x^2}=16x^7 ➖ (76)^2= 5776 {16x^7 ➖ 5776}= 5760x^7 3^19^6^10x^7 3^19^1^6^2^5x^7 3^1^1^3^2^1^1x^1 1^1^3^2x 3^2x(x ➖ 3x+2).