Master Radical Math with This Simple Trick! | Binomial Expansion

[@SidneiMV]

nice shortcut at the end

[@潘博宇-k4l]

E=X^6=[148-55(7)^(1/2)]/512.

[@Shobhamaths]

E=(148-55√7) /512

[@Quest3669]

X=( √7-1))/4 hence
? = (148-55√7)/512 soln.

[@guyhoghton399]

Here is another shortcut(?) to find _x⁶:_
_x = ¼(√7 - 1)_
Let *_p = √7 + 1, q = √7 - 1_*
_p - q = 2_
_pq = (√7)² - 1² = 6_
_(p - q)² = 4 = p² + q² - 2pq_
⇒ _p² + q² = 4 + 2(6) = 16_
_(p² + q²)³ = (2⁴)³ = p⁶ + q⁶ + (3)(pq)²(p² + q²)_
⇒ _p⁶ + q⁶ = 2¹² - (3)(6²)(16) = 4096 - 1728 = 2368_
Let *_(t - p⁶)(t - q⁶) = 0_* ... ①
⇒ t² - (p⁶ + q⁶)t + (pq)⁶ = 0
⇒ _t² - 2368t + 6⁶ = 0_
⇒ _t = 1184 ± √(1184² - 46656) = 1184 ± √(1,355,200) = 1184 ± √(2⁶.5².7.11²)_
⇒ _p⁶, q⁶ = 1184 ± (2³)(5)(11)√7_ by inspection with ①
⇒ _q⁶ = 1184 - 440√7_ since _|q| < |p|_
∴ *_x⁶ = (¼q)⁶ = (1184 - 440√7)/8⁴ = (184 - 55√7)/512_*

[@SidneiMV]

84 = 2²3¹7¹ = 12¹7¹
36 = 2²3¹3¹ = 12¹3¹
28 = 2²7¹ = 4¹7¹
12 = 2²3¹ = 4¹3¹
1/x = [√12(√7 + √3) + √4(√7 + 3)]/[(√12 + √4) + (√7 + √3)]
1/x = (√12 + √4)(√7 + √3)/[(√12 + √4) + (√7 + √3)]
x = 1/(√7 + √3) + 1/(√12 + √4)
x = (√7 - √3)/4 + (√12 - √4)/8
x = (2√7 - 2√3 + 2√3 - 2)/8
x = (√7 - 1)/4
*x⁶ = [(√7 - 1)/4]⁶*