An Interesting Cubic Equation | Algebra Insights

[@kassuskassus6263]

One real solution x=0. Let x+5=t and we have to solve for t after simplifications, 2t^3-3t^2+33t-340=0. We'll get one real solution which is t=5, so x=0.

[@RashmiRay-c1y]

Setting x+9/2 =0, we get 4t^3+63t - 648=0, which has t=9/2 as the only real solution. Thus, x=0.

[@nasrullahhusnan2289]

Note that 1³+2³+...+n²=[½n(n+1)]² So 1³+2³+...+8³=(½×8×9)² =36² =6⁴ Thus x=0

[@mohammedsaysrashid3587]

It was a wonderful introduction....thanks Sir 🙏 for sharing x= (u-9)/2=>0 ...x=0

[@mikelivstone]

This can be solved directly using the formula you finished writing at 4:37. Just subtract the sum of the first x cubes from the sum of the first x+8 cubes.

[@Quest3669]

X= 0

[@SidneiMV]

x + 4 = u/2 - 1/2 => x = (u - 9)/2 (u - 7)³ + (u - 5)³ + (u - 3)³ + (u - 1)³ + (u + 1)³ + (u + 3)³ + (u + 5)³ + (u + 7)³ = 2⁷3⁴ 2(u³ + (3)7²u) + 2(u³ + (3)5²u) + 2(u³ + (3)3²u) + 2(u³ + (3)1²u) = 2⁷3⁴ 8u³ + (6)(1² + 3² + 5² + 7²)u - 2⁷3⁴ = 0 8u³ + (6)84u - 2⁷3⁴ = 0 u³ + 63u - 1296 = 0 u = 9 (by inspection) x = (u - 9)/2 => *x = 0*

[@АндрейПергаев-з4н]

Если сделать замену, х+4,5=а, (среднее арифметическое оснований степени) то при раскрытии кубов две пары слагаемых сократятся Выражение вида (x+y)^3+(x-y)^3 получается в итоге

[@Fjfurufjdfjd]

χ=0 στο R

[@manasishastry345]

let x+4=a then solve .it is very simple calculation.

[@RealQinnMalloryu4]

{3x^3+8x3}= 11x^6{ 9x^3+15x^3}=24x^6.{11x^6+24x^6}={35x^12 +24x^3 }= 59x^15 3^17x^15 3^17^1x^3^5 1^1^1x^3^5^1 x^3^1^1 x^3^1 (x ➖ 3x+1)

[@RealQinnMalloryu4]

6^4 3^22^2 3^1^1^1 3^1 (x ➖ 3x+1)