@@haiderlughmani BUT THIS GUY SAYS THAT YOU MESSED UP AT DIFFERENT LEVELS. The mistake comes from earlier mistake. At 12:48 it is written 9x⁵=-5x+2. But upon recalling that x=[cbrt(10)-1]/3 it was the written 9x⁵=-5[{cbrt(10)-2}/3]⁵+2 in 13:21. Supposedly 9x⁵=-5[{cbrt(10)-1}/3]+2 Mistake also occurs at 13:57 when opening bracket. It must be 9x⁵=[6-5-5cbrt(10)]/3
The mistake comes from earlier mistake. At 12:48 it is written 9x⁵=-5x+2. But upon recalling that x=[cbrt(10)-1]/3 it was the written 9x⁵=-5[{cbrt(10)-2}/3]⁵+2 in 13:21. Supposedly 9x⁵=-5[{cbrt(10)-1}/3]+2 Mistake also occurs at 13:57 when opening bracket. It must be 9x⁵=[6-5-5cbrt(10)]/3
Re-post this problem after rectifying the mistakes raised by ALL.
At last it will be 16 not 11. its Miner Mistake.
@@haiderlughmani
BUT THIS GUY SAYS THAT YOU MESSED UP AT DIFFERENT LEVELS.
The mistake comes from earlier mistake. At 12:48 it is written 9x⁵=-5x+2. But upon recalling that x=[cbrt(10)-1]/3 it was the written 9x⁵=-5[{cbrt(10)-2}/3]⁵+2 in 13:21. Supposedly
9x⁵=-5[{cbrt(10)-1}/3]+2
Mistake also occurs at 13:57 when opening bracket. It must be
9x⁵=[6-5-5cbrt(10)]/3
@@haiderlughmani man don't forget. I am a sincere follower.
❤❤❤
au final c'est [ 16 - 5 * (10)^(1/3) ] /27 ===> 16 au lieu de 11
Excellente démonstration
Let a=[(10^⅓)-1]⁵
Expanding
a=(10^⅓)⁵-5(10^⅓)⁴+10(10^⅓)³
-10(10^⅓)² +5(10^⅓)+1
=10(10^⅓)²-50(10^⅓)+100
-10(10^⅓)²-5(10^⅓)+1
=101-55(10^⅓)
=56+55[1-(10^⅓)]
=56-55a
56a=56 ---> a=1
Thus [⅓{(10^⅓)-1]⁵=(⅓)⁵
there is a mistake in the final result
Sory its 16 not 11
Así es... rectificar es de sabios 😊
6+10=16, not 11.
The mistake comes from earlier mistake. At 12:48 it is written 9x⁵=-5x+2. But upon recalling that x=[cbrt(10)-1]/3 it was the written 9x⁵=-5[{cbrt(10)-2}/3]⁵+2 in 13:21. Supposedly
9x⁵=-5[{cbrt(10)-1}/3]+2
Mistake also occurs at 13:57 when opening bracket. It must be
9x⁵=[6-5-5cbrt(10)]/3
Excelente resolução mas no final quando vai substituir o valor de X você esqueceu do -1 e continuou a resolução.