If 1/a+1/b+1/c =1/(a+b+c) then prove that 1/a^3+1/b^3+1/b^3 =1/(a^3+b^3+c^3)=1/(a+b+c)^3
If 1/a+1/b+1/c =1/(a+b+c) then prove that 1/a^3+1/b^3+1/b^3 =1/(a^3+b^3+c^3)=1/(a+b+c)^3
Md Mursalim Saikh , 7 Years ago
Grade 10
Algebra> If 1/a+1/b+1/c =1/(a+b+c) then prove that...
1 Answers
Arun
Write u = 1/a, v = 1/b, w = 1/c. Then 1/(a+b+c) = 1/(1/u + 1/v + 1/w) = uvw/(vw + uw + uv).
By assumption u + v + w = uvw/(uv + uw + vw) ==>
uvw = (u+v+w)(uv+uw+vw) = u^2v + u^2w + uvw + uv^2 + uvw + v^2w + uvw + uw^2 + vw^2 ==>
uvw = 3uvw + u^2v+ u^2w + uv^2 + v^2w + uw^2 + vw^2 ==>
-2uvw = u^2v+ u^2w + uv^2 + v^2w + uw^2 + vw^2 ==>
(u+v)w^2 + (u^2 + v^2 + 2uv)w +(u^2v+uv^2) = 0 ==>
w = {-(u+v)^2+/-sqrt((u+v)^4 - 4uv(u+v)^2)}/[2(u+v)] = {-(u+v) +/-(u-v)}/2 = -v or - u.
Going back to u = 1/a, v = 1/b and w = 1/c, then w = - v or -u means c = - a or - b.
So 1/a + 1/b + 1/c = 1/(a+b+c) ==> c = -a or - b ==>
c^3 = -a^3 or - b^3 ==>
1/a^3 + 1/b^3 + 1/c^3 = 1/(a^3 + b^3 + c^3) {since the c^3 term will cancel either the a^3 or b^3 term on both sides, and you'll be left with say 1/a^3 = 1/a^3}, and we're done.
By assumption u + v + w = uvw/(uv + uw + vw) ==>
uvw = (u+v+w)(uv+uw+vw) = u^2v + u^2w + uvw + uv^2 + uvw + v^2w + uvw + uw^2 + vw^2 ==>
uvw = 3uvw + u^2v+ u^2w + uv^2 + v^2w + uw^2 + vw^2 ==>
-2uvw = u^2v+ u^2w + uv^2 + v^2w + uw^2 + vw^2 ==>
(u+v)w^2 + (u^2 + v^2 + 2uv)w +(u^2v+uv^2) = 0 ==>
w = {-(u+v)^2+/-sqrt((u+v)^4 - 4uv(u+v)^2)}/[2(u+v)] = {-(u+v) +/-(u-v)}/2 = -v or - u.
Going back to u = 1/a, v = 1/b and w = 1/c, then w = - v or -u means c = - a or - b.
So 1/a + 1/b + 1/c = 1/(a+b+c) ==> c = -a or - b ==>
c^3 = -a^3 or - b^3 ==>
1/a^3 + 1/b^3 + 1/c^3 = 1/(a^3 + b^3 + c^3) {since the c^3 term will cancel either the a^3 or b^3 term on both sides, and you'll be left with say 1/a^3 = 1/a^3}, and we're done.
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