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If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=?
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If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0...
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If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0...
Problem:
If a, b, c are real numbers (not equal to 0), and a/bc + b/ac + c/ab = 1, then find the value of (ab)(bc)(ca).
Solution:
Given:
a/bc + b/ac + c/ab = 1
We need to find the value of (ab)(bc)(ca).
Approach:
To find the value of (ab)(bc)(ca), we can simplify the given equation and relate it to the expression we need to find.
Simplifying the given equation:
Multiplying the given equation by abc, we get:
a^2 + b^2 + c^2 = abc
Using the identity:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
Substituting the value of a^2 + b^2 + c^2 from the simplified equation:
(a + b + c)^2 = abc + 2(ab + bc + ca)
Simplifying further:
(a + b + c)^2 = 2(ab + bc + ca) + abc
Expanding (a + b + c)^2:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
Expanding the left-hand side:
a^2 + b^2 + c^2 + 2(ab + bc + ca) = 2(ab + bc + ca) + abc
Cancelling out terms:
a^2 + b^2 + c^2 = abc
This equation is the same as the one we obtained earlier by simplifying the given equation.
Using the equation to find the value of (ab)(bc)(ca):
Since a^2 + b^2 + c^2 = abc, we can rewrite it as:
(ab)(bc) + (bc)(ca) + (ca)(ab) = (abc)^2
Multiplying both sides by abc, we get:
(ab)(bc)(ca) + (bc)(ca)(ab) + (ca)(ab)(bc) = (abc)^3
Answer:
Therefore, the value of (ab)(bc)(ca) is (abc)^3.
If a, b, c are real numbers (not equal to 0), and a/bc + b/ac + c/ab = 1, then find the value of (ab)(bc)(ca).
Solution:
Given:
a/bc + b/ac + c/ab = 1
We need to find the value of (ab)(bc)(ca).
Approach:
To find the value of (ab)(bc)(ca), we can simplify the given equation and relate it to the expression we need to find.
Simplifying the given equation:
Multiplying the given equation by abc, we get:
a^2 + b^2 + c^2 = abc
Using the identity:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
Substituting the value of a^2 + b^2 + c^2 from the simplified equation:
(a + b + c)^2 = abc + 2(ab + bc + ca)
Simplifying further:
(a + b + c)^2 = 2(ab + bc + ca) + abc
Expanding (a + b + c)^2:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
Expanding the left-hand side:
a^2 + b^2 + c^2 + 2(ab + bc + ca) = 2(ab + bc + ca) + abc
Cancelling out terms:
a^2 + b^2 + c^2 = abc
This equation is the same as the one we obtained earlier by simplifying the given equation.
Using the equation to find the value of (ab)(bc)(ca):
Since a^2 + b^2 + c^2 = abc, we can rewrite it as:
(ab)(bc) + (bc)(ca) + (ca)(ab) = (abc)^2
Multiplying both sides by abc, we get:
(ab)(bc)(ca) + (bc)(ca)(ab) + (ca)(ab)(bc) = (abc)^3
Answer:
Therefore, the value of (ab)(bc)(ca) is (abc)^3.
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If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=?
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If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=?.
If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a,b,c are real no. a,b,c not equal to 0 and a+b+c is not equal to 0 and 1/a+1/b+1/x=1/a+b+c then (a+b)(b+c)(c+a)=?.
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