Can you solve This If-Then Algebraic Question? | Step-by-Step Explanation

Learn how to solve this If-Then question:
If a + b = 2 and 1/a + 1/b =2
Then a^3 + b^3 = ?

[geoffreyparfitt]

From inspection a = b = 1 and the symmetry of the system of equations establishes this as the only solution. No algebra needed.

[botfeeder]

It's an elegant solution but just solving individually for a and b seems to be just as straightforward. From equation 1, a=2-b. Substitute that into the second equation and juggle that equation around to get b^2-2b+1=0. Or (b-1)*(b-1)=0. Thus b=1. By equation 1 that says that a=1 hence a^3+b^3= 1+1=2.

[tambuwalmathsclass]

I really enjoyed watching your contents, keep on the good work 👍

[aminefan]

We love your explications, Moroccan people love you

[lavc]

Dear teacher, your questions gives us a chance to think before we start our day .

[davidfromstow]

First one I've managed to solve - mainly by logic/instinct rather than your elegant solution!

[drpkmath]

Love the system of equations my friend! Great problem and nice presentation! Keep rocking!

[math]

sure can, thanks for sharing bro, love the challenge

[paulc]

Thank you sir. After watched your prevoious similar videos, I know how to work it out easily.

[JohnRandomness]

Trivial. 2 = 1/a + 1/b = (b+a)/ab = 2/ab. So ab = 1. Therefore a + 1/a = 2. Two is the minimum value of a + 1/a, and appears when a = 1. Therefore a = b = 1 and a³ + b³ = 2.

[GetMeThere]

By inspection, it's obvious that a=1, b=1 is A solution. What I'm wondering is whether having "A" solution is enough to give a "proper" answer to the question. Are there other solutions to the first two equations, where a^3 + b^3 might NOT equal 1?

[satyamkumar]

Sir I got it like this
If both are equal to 2, so they itself should be equal:
a+b = 1/a + 1/b
By solving we got a.b = 1
Therefore b = 1/a
By substituting it in a + b = 2
We got a quadratic eq and by solving it we got value of 'a' as 1
Again by substituting 1 in a+b=2
We got 'b' as 1
Hence, 1 cube + 1 cube = 2
That's my final ans.

[devondevon]

obviously a^3+b^3=2
since a=1 and b=1 satisfy both a+b=2, AND 1/a + 1/b=2
I found ab differently.
below is how I solve it:
(1/a + 1/b)(a+b) =4 (multiply equation 1 and 2)
a/b +b/a + 2 =4
a/b + b/a =2 equation 3
b^2 + a^2 = 2ab (multiply both sides by 2ab) to get equation 4

(a +b)^2= 4 (square equation 1)
a^2 + b^2 + 2ab= 4 to get equation 5
2ab + 2ab =4 (substitute equation 4 into equation 5)
4ab = 4
ab =1
since a^3 + b^3= (a+b)(a^2 + b^2-ab)
then a^3+b^3 = 2 (2ab-ab)
a^3 +b^3 =2ab
but ab=1, so a^3 + b^3 =2(1) =2
Answer =2 4:23

[piman]

Isn't a = 1, b = 1 the obvious (and only) solution - thus leading to a^3 + b^3 = 2?

[lzuluaga]

Elevé a+b al cubo, factoricé los dos términos centrales he hice los reemplazos.

[lialos]

Rather than your method, in the middle, I went to (a+b)^3=a^3+3a^b+3ab^2+b^3, which is also equal to 8. And then looking at the middle two terms as 3a^2b+3ab^2=3ab(a+b), we know ab=1 (from what you started with), and a+b=2(from what was given), so 3ab(a+b)=6. That gives a^3+6+b^3=8, so a^3+b^3=2.

[MyOneFiftiethOfADollar]

Squaring and cubing unnecessary. ab=1 means a and b are reciprocals. Only way the sum
Of two reciprocals equals 2 is a=b=1 implies a^3 + b^3=2

[pratapjadhao]

Thanks sir
(Put a =b-2
In second eq and solve)

[philipkudrna]

a=b=1, thus the solution is 2. Took me a small piece of paper and some algebra… I put b = 2-a into the second equation and then ended up with (a-1)^2 to solve…

[mustafizrahman]

a+b=2.Needed 10 seconds. I have easily solved it.