Answer
397.5k+ views
Hint:
Here, we will use the given condition to find one variable in terms of the other two variables. Then we will take each expression of the equation and substitute the obtained variable in each expression. We will simplify it further and prove that the given statement is true.
Formula Used:
The square of the sum of the numbers is given by an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Complete step by step solution:
The given equation is \[a + b + c = 0\].
Now, by rewriting the equation, we get
\[ \Rightarrow c = - \left( {a + b} \right)\] ………………………………………………………………….\[\left( 1 \right)\]
Now, we will prove the given condition \[{a^2} - bc = {b^2} - ca = {c^2} - ab\] one by one and we will prove that all the three given conditions leads to the same result.
First, we will find the result of the first expression, \[{a^2} - bc\], by substituting equation \[\left( 1 \right)\]. Therefore, we get
\[{a^2} - bc = {a^2} - b\left( { - \left( {a + b} \right)} \right)\]
We know that the product of two negative integers is a positive integer, so we get
\[ \Rightarrow {a^2} - bc = {a^2} + b\left( {a + b} \right)\]
By multiplying the terms, we get
\[ \Rightarrow {a^2} - bc = {a^2} + {b^2} + ab\] ………………………………………………….\[\left( 2 \right)\]
Now, we will find the result of the second expression \[{b^2} - ca\] by substituting equation \[\left( 1 \right)\].
\[{b^2} - ca = {b^2} - \left( { - \left( {a + b} \right)} \right)a\]
We know that the product of two negative integers is a positive integer, so we get
\[ \Rightarrow {b^2} - ca = {b^2} + \left( {a + b} \right)a\]
Multiplying the terms, we get
\[ \Rightarrow {b^2} - ca = {b^2} + {a^2} + ab\] ………………………………………………….\[\left( 3 \right)\]
Now, we will find the result of the third expression \[{c^2} - ca\] by substituting equation\[\left( 1 \right)\].
\[{c^2} - ca = {\left( { - \left( {a + b} \right)} \right)^2} - ca\]
We know that the product of two negative integers is a positive integer.
Now, by using an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow {c^2} - ab = {a^2} + {b^2} + 2ab - ab\]
Simplifying the expression, we get
\[ \Rightarrow {c^2} - ab = {a^2} + {b^2} + ab\] ………………………………………………….\[\left( 4 \right)\]
So, we get \[{a^2} - bc = {b^2} - ca = {c^2} - ab = {a^2} + {b^2} + ab\]
Therefore, \[{a^2} - bc = {b^2} - ca = {c^2} - ab\] is proved true when \[a + b + c = 0\].
Note:
We know that the given expression is an algebraic expression. An algebraic expression is defined as an expression with the combination of variables, constants and operators. We can also find the variables \[a\] and \[b\] in terms of the other variable by using the given condition. So, we get \[a = - \left( {b + c} \right)\] and \[b = - \left( {a + c} \right)\]. By substituting these variables in the given expression we will prove the same results which are equal for all the three expressions.
Here, we will use the given condition to find one variable in terms of the other two variables. Then we will take each expression of the equation and substitute the obtained variable in each expression. We will simplify it further and prove that the given statement is true.
Formula Used:
The square of the sum of the numbers is given by an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Complete step by step solution:
The given equation is \[a + b + c = 0\].
Now, by rewriting the equation, we get
\[ \Rightarrow c = - \left( {a + b} \right)\] ………………………………………………………………….\[\left( 1 \right)\]
Now, we will prove the given condition \[{a^2} - bc = {b^2} - ca = {c^2} - ab\] one by one and we will prove that all the three given conditions leads to the same result.
First, we will find the result of the first expression, \[{a^2} - bc\], by substituting equation \[\left( 1 \right)\]. Therefore, we get
\[{a^2} - bc = {a^2} - b\left( { - \left( {a + b} \right)} \right)\]
We know that the product of two negative integers is a positive integer, so we get
\[ \Rightarrow {a^2} - bc = {a^2} + b\left( {a + b} \right)\]
By multiplying the terms, we get
\[ \Rightarrow {a^2} - bc = {a^2} + {b^2} + ab\] ………………………………………………….\[\left( 2 \right)\]
Now, we will find the result of the second expression \[{b^2} - ca\] by substituting equation \[\left( 1 \right)\].
\[{b^2} - ca = {b^2} - \left( { - \left( {a + b} \right)} \right)a\]
We know that the product of two negative integers is a positive integer, so we get
\[ \Rightarrow {b^2} - ca = {b^2} + \left( {a + b} \right)a\]
Multiplying the terms, we get
\[ \Rightarrow {b^2} - ca = {b^2} + {a^2} + ab\] ………………………………………………….\[\left( 3 \right)\]
Now, we will find the result of the third expression \[{c^2} - ca\] by substituting equation\[\left( 1 \right)\].
\[{c^2} - ca = {\left( { - \left( {a + b} \right)} \right)^2} - ca\]
We know that the product of two negative integers is a positive integer.
Now, by using an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow {c^2} - ab = {a^2} + {b^2} + 2ab - ab\]
Simplifying the expression, we get
\[ \Rightarrow {c^2} - ab = {a^2} + {b^2} + ab\] ………………………………………………….\[\left( 4 \right)\]
So, we get \[{a^2} - bc = {b^2} - ca = {c^2} - ab = {a^2} + {b^2} + ab\]
Therefore, \[{a^2} - bc = {b^2} - ca = {c^2} - ab\] is proved true when \[a + b + c = 0\].
Note:
We know that the given expression is an algebraic expression. An algebraic expression is defined as an expression with the combination of variables, constants and operators. We can also find the variables \[a\] and \[b\] in terms of the other variable by using the given condition. So, we get \[a = - \left( {b + c} \right)\] and \[b = - \left( {a + c} \right)\]. By substituting these variables in the given expression we will prove the same results which are equal for all the three expressions.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)