Math, asked by mrjatt69, 4 days ago

if a not equal to b then show that (a+b) [1/a+1/b]​

Answers

Answered by sonakshi1jadhav
2

Answer:

a^3=b^3.

Step-by-step explanation:

If 1/a+1/b=1/(a+b), prove that a^3=b^3. Is this question solvable? If yes, how?

Implicit in the statement of the equation 1a+1b=1a+b is the assumption that a≠0, b≠0, and a+b≠0. What is not clear is the domain where a and b lie in.

The given equation reduces to (a+b)2=ab, and hence to a2+ab+b2=0. Thus t2+t+1=0, where t=ab. Each solution to t2+t+1=0 is also a solution to (t−1)(t2+t+1)=0, or to t3–1=0. Therefore a3=(bt)3=b3. ■

How do I prove that 13+23+33+43+⋅⋅⋅+n3 is equal to (1+2+3+4+⋅⋅⋅+n)2?

If x+1/x=3, what is the value of x3+1/x3=?

How do you prove that 3 (a^3+b^3 + c^3) >= (ab + bc + AC)?

If abc=1 and a,b,c>0, then how can I prove that 1a3(c+b)+1b3(c+a)+1c3(b+a)≥32?

What is 1+1-3+1-3 (1-3(1-3))?

Yes, it is solvable.

The given expression is 1/a+1/b=1/(a+b)

Therefore, by taking L. C. M on the left hand side we get,

(a+b) /ab=1/(a+b)

Now, by cross multiplication we get,

(a+b) ^2=ab

Thus,a^2+b^2+2ab=ab

Hence, a^2+b^2+2ab-ab=0

a^2+b^2+ab=0…………eqn(1)

We know, a^3-b^3=(a-b)(a^2+b^2+ab)

From eqn(1).. we get,

a^3-b^3=(a-b)*0

Hence, a^3-b^3=0.

Therefore, a^3=b^3.

HENCE PROVED.

HOPE U LIKE THE ANSWER.

THANKYOU…….

Answered by kanchankumari0201198
1

Step-by-step explanation:

(a+b)= [1/a+1/b]

then

answer will be

a+b=2/ab

(a+b) ab=2

Similar questions