Prove $frac{a^2b^2}{(a-c)(b-c)}+ frac{b^2c^2}{(b-a)(c-a)} + frac{c^2a^2}{(c-b)(a-b)} = ab+bc+ca$
$begingroup$
Prove $$frac{a^2b^2}{(a-c)(b-c)}+ frac{b^2c^2}{(b-a)(c-a)} + frac{c^2a^2}{(c-b)(a-b)} = ab+bc+ca$$
Can you help me to prove this equality? When I multiple left side, I get more complicated expression. Obviously I need to add something like shift for our variables or something else. I would be grateful for help.
algebra-precalculus
edited Jan 22 at 21:08
StubbornAtom
6,05811239
asked Jan 22 at 19:11
bubblesPbubblesP
122
$endgroup$
add a comment |
$begingroup$
Prove $$frac{a^2b^2}{(a-c)(b-c)}+ frac{b^2c^2}{(b-a)(c-a)} + frac{c^2a^2}{(c-b)(a-b)} = ab+bc+ca$$
Can you help me to prove this equality? When I multiple left side, I get more complicated expression. Obviously I need to add something like shift for our variables or something else. I would be grateful for help.
algebra-precalculus
edited Jan 22 at 21:08
StubbornAtom
6,05811239
asked Jan 22 at 19:11
bubblesPbubblesP
122
$endgroup$
2
$begingroup$
For $a=b=c$ the RHS is $3a^2$ and the LHS is $infty$.
$endgroup$
– Dietrich Burde
Jan 22 at 19:13
$begingroup$
The Statement is true, we must assume that $$ane c,bne c,ane b$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:19
add a comment |
$begingroup$
Prove $$frac{a^2b^2}{(a-c)(b-c)}+ frac{b^2c^2}{(b-a)(c-a)} + frac{c^2a^2}{(c-b)(a-b)} = ab+bc+ca$$
Can you help me to prove this equality? When I multiple left side, I get more complicated expression. Obviously I need to add something like shift for our variables or something else. I would be grateful for help.
algebra-precalculus
edited Jan 22 at 21:08
StubbornAtom
6,05811239
asked Jan 22 at 19:11
bubblesPbubblesP
122
$endgroup$
Prove $$frac{a^2b^2}{(a-c)(b-c)}+ frac{b^2c^2}{(b-a)(c-a)} + frac{c^2a^2}{(c-b)(a-b)} = ab+bc+ca$$
Can you help me to prove this equality? When I multiple left side, I get more complicated expression. Obviously I need to add something like shift for our variables or something else. I would be grateful for help.
algebra-precalculus
algebra-precalculus
edited Jan 22 at 21:08
StubbornAtom
6,05811239
asked Jan 22 at 19:11
bubblesPbubblesP
122
edited Jan 22 at 21:08
StubbornAtom
6,05811239
asked Jan 22 at 19:11
bubblesPbubblesP
122
edited Jan 22 at 21:08
StubbornAtom
6,05811239
edited Jan 22 at 21:08
StubbornAtom
6,05811239
edited Jan 22 at 21:08
StubbornAtom
6,05811239
6,05811239
asked Jan 22 at 19:11
bubblesPbubblesP
122
asked Jan 22 at 19:11
bubblesPbubblesP
122
asked Jan 22 at 19:11
bubblesPbubblesP
122
122
2
$begingroup$
For $a=b=c$ the RHS is $3a^2$ and the LHS is $infty$.
$endgroup$
– Dietrich Burde
Jan 22 at 19:13
$begingroup$
The Statement is true, we must assume that $$ane c,bne c,ane b$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:19
add a comment |
2
$begingroup$
For $a=b=c$ the RHS is $3a^2$ and the LHS is $infty$.
$endgroup$
– Dietrich Burde
Jan 22 at 19:13
$begingroup$
The Statement is true, we must assume that $$ane c,bne c,ane b$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:19
2
2
$begingroup$
For $a=b=c$ the RHS is $3a^2$ and the LHS is $infty$.
$endgroup$
– Dietrich Burde
Jan 22 at 19:13
$begingroup$
For $a=b=c$ the RHS is $3a^2$ and the LHS is $infty$.
$endgroup$
– Dietrich Burde
Jan 22 at 19:13
$begingroup$
The Statement is true, we must assume that $$ane c,bne c,ane b$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:19
$begingroup$
The Statement is true, we must assume that $$ane c,bne c,ane b$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Multiply the whole thing by the product of the denominators.
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
$endgroup$
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
add a comment |
$begingroup$
Idea:
If we clear the denominators we get $$ a^2b^2(a-b) + b^2c^2(b-c)+c^2a^2(c-a)=(ab+bc+ca)(a-b)(b-c)(a-c)$$
which are polynomials of degree 3 (say on $a$) on both sides. So the identity is true if it is true of 4 different values of $a$
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Multiply the whole thing by the product of the denominators.
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
$endgroup$
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
add a comment |
$begingroup$
Hint: Multiply the whole thing by the product of the denominators.
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
$endgroup$
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
add a comment |
$begingroup$
Hint: Multiply the whole thing by the product of the denominators.
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
$endgroup$
Hint: Multiply the whole thing by the product of the denominators.
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
answered Jan 22 at 19:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
76.6k42866
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
add a comment |
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
$begingroup$
I solved this. Thank you a lot people.
$endgroup$
– bubblesP
Jan 22 at 19:36
add a comment |
$begingroup$
Idea:
If we clear the denominators we get $$ a^2b^2(a-b) + b^2c^2(b-c)+c^2a^2(c-a)=(ab+bc+ca)(a-b)(b-c)(a-c)$$
which are polynomials of degree 3 (say on $a$) on both sides. So the identity is true if it is true of 4 different values of $a$
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
$endgroup$
add a comment |
$begingroup$
Idea:
If we clear the denominators we get $$ a^2b^2(a-b) + b^2c^2(b-c)+c^2a^2(c-a)=(ab+bc+ca)(a-b)(b-c)(a-c)$$
which are polynomials of degree 3 (say on $a$) on both sides. So the identity is true if it is true of 4 different values of $a$
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
$endgroup$
add a comment |
$begingroup$
Idea:
If we clear the denominators we get $$ a^2b^2(a-b) + b^2c^2(b-c)+c^2a^2(c-a)=(ab+bc+ca)(a-b)(b-c)(a-c)$$
which are polynomials of degree 3 (say on $a$) on both sides. So the identity is true if it is true of 4 different values of $a$
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
$endgroup$
Idea:
If we clear the denominators we get $$ a^2b^2(a-b) + b^2c^2(b-c)+c^2a^2(c-a)=(ab+bc+ca)(a-b)(b-c)(a-c)$$
which are polynomials of degree 3 (say on $a$) on both sides. So the identity is true if it is true of 4 different values of $a$
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
answered Jan 22 at 19:28
greedoidgreedoid
45k1157112
45k1157112
add a comment |
add a comment |
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2
$begingroup$
For $a=b=c$ the RHS is $3a^2$ and the LHS is $infty$.
$endgroup$
– Dietrich Burde
Jan 22 at 19:13
$begingroup$
The Statement is true, we must assume that $$ane c,bne c,ane b$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:19