What an incredibly cumbersome solution! Just expand the original expression with the binomial theorem, do some simple arithmetic, collect like terms, and the result drops out with about four lines of calvulation
@@TonyFisher-lo8hh Agreed, this is the general approach to handle such expression. By the way: to correct typos, press on the 3 points that are located right of you text and simply change your text. ('calculation' in this case)
Podemos resolver também por produtos notáveis(quadrado da diferença), separando em 3 parcelas (V2 - 1)'2. Depois, teremos novamente 3 parcelas iguais e podemos resolver duas delas por produto notável(quadrado da diferença). E o resultado multiplica com o termo que sobrou.
I watched some of your videos. You keep solving the same trivial (but you call it "HARD" for some reason) algebra problem over and over again, just with different numbers. What could be the purpose of this?
why not use pascals triangle to figure out the formula for (a-b)^6 and solving it??
Or at least (a + sqrt(b)) ^ n ...
@@rainerzufall42 yea thats what i was sayin
1 * 8 + 15 * 4 + 15 * 2 + 1 * 1 = 8 + 60 + 30 + 1 = 99
6 * 4 + 20 * 2 + 6 * 1 = 24 + 40 + 6 = 70
=> (sqrt(2) - 1)^6 = 99 - 70 sqrt(2). Easy.
Without any calculation: This should be 1 / (99 + 70 sqrt(2)) ~= 1 / 198 = 0.0050505... (good enough!).
Why that? (sqrt(2) - 1) = 1 / (sqrt(2) + 1) and 99 ~= 70 sqrt(2), because |sqrt(2) - 1|
You are real teacher.
(rt2-1)^2=3-2rt2
(rt2-1)^4=17-12rt2
(rt2-1)2×(rt2-1)^4
=(3-2rt2)(17-12rt2)
=51+48-70rt2
=99-70rt2
What an incredibly cumbersome solution! Just expand the original expression with the binomial theorem, do some simple arithmetic, collect like terms, and the result drops out with about four lines of calvulation
Sorry: calculation
@@TonyFisher-lo8hh Agreed, this is the general approach to handle such expression. By the way: to correct typos, press on the 3 points that are located right of you text and simply change your text. ('calculation' in this case)
@@ralkadde "Three dots" to enter edit-mode works in Chrome under Windows, but not in youtube under Android.
@@TonyFisher-lo8hh Thank you for this information. I did not know.
Thank you very much
Podemos resolver também por produtos notáveis(quadrado da diferença), separando em 3 parcelas (V2 - 1)'2. Depois, teremos novamente 3 parcelas iguais e podemos resolver duas delas por produto notável(quadrado da diferença). E o resultado multiplica com o termo que sobrou.
Just find cube and get result and then multiply with same result .. within 20 seconds 😂
С помощью дискриминала решили бы давно.
(2 ➖ 1)^2^3 (1 ➖ 1)^2^3 (x ➖ 3x+2).
❤❤
I watched some of your videos. You keep solving the same trivial (but you call it "HARD" for some reason) algebra problem over and over again, just with different numbers. What could be the purpose of this?
Its simpler to square and then cube!!No trick ,much easier.
With in 30 sec answer aa Jayega
Please 🙏 tell kaise aayega , Please 🙏
Binomial theorem se Karo 😅😅
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I love Your types of Mathematics Please Pin my comment🎉
Are you fun
I love Your types of Mathematics Please Pin my comment🎉