Question

prove that \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\] . How can I solve this without expanding everything?

Answer 1

Hint: Here in this question, we have to prove right hand side by solving the left hand side (i.e., Prove LHS=RHS) this LHS can be solve by using the substitution of algebraic identities and simplify using the basic arithmetic operation to get the required value of RHS.

Complete step by step solution:
The equation is in the form of the algebraic equation, where the algebraic equation is a combination of variables and the constants.
Now consider the given equation
 \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\] --------(1)
As we know the cubic algebraic identity
 \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\]
On rearranging the above algebraic identity can be written as
 \[ \Rightarrow\,\,{a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)\] --------(2)
Take \[\left( {a + b} \right) = d\] , then
Equation (2) becomes
 \[ \Rightarrow\,\,{a^3} + {b^3} = {d^3} - 3abd\] ---------(3)
Now, consider the LHS of equation (1)
 \[ \Rightarrow \,LHS\]
 \[ \Rightarrow \,{a^3} + {b^3} + {c^3} - 3abc\]
Substitute equation (3) in the above equation we get
 \[ \Rightarrow{d^3} - 3abd + {c^3} - 3abc\]
On rearranging the above equation we get
 \[ \Rightarrow{d^3} + {c^3} - 3abd - 3abc\]
Taking the common terms we get
 \[ \Rightarrow{d^3} + {c^3} - 3ab\left( {d + c} \right)\]
Add and subtract \[3dc\left( {d + c} \right)\] to above equation
 \[ \Rightarrow{d^3} + {c^3} + 3dc\left( {d + c} \right) - 3dc\left( {d + c} \right) - 3ab\left( {d + c} \right)\]
Again, by algebraic identity \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\] can be written as
 \[ \Rightarrow{\left( {d + c} \right)^3} - 3dc\left( {d + c} \right) - 3ab\left( {d + c} \right)\] ----------(4)
Take \[\left( {d + c} \right)\] as common and the equation is written as
 \[ \Rightarrow\left( {d + c} \right)\left\{ {{{\left( {d + c} \right)}^2} - 3dc - 3ab} \right\}\]
Now use the standard algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] . Then we have
 \[ \Rightarrow\left( {d + c} \right)\left\{ {{d^2} + {c^2} + 2dc - 3dc - 3ab} \right\}\]
On simplifying the above equation, we get
 \[ \Rightarrow\left( {d + c} \right)\left\{ {{d^2} + {c^2} - dc - 3ab} \right\}\]
Now, put \[d = a + b\] , On substituting the equation is written as
 \[ \Rightarrow\left( {a + b + c} \right)\left\{ {{{\left( {a + b} \right)}^2} + {c^2} - \left( {a + b} \right)c - 3ab} \right\}\]
Again, by the identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
 \[ \Rightarrow\left( {a + b + c} \right)\left\{ {{a^2} + {b^2} + 2ab + {c^2} - ac - bc - 3ab} \right\}\]
On simplification, we get
 \[ \Rightarrow\left( {a + b + c} \right)\left\{ {{a^2} + {b^2} + {c^2} - ac - bc - ab} \right\}\]
 \[ \Rightarrow \,RHS\]
Hence we have proved the LHS is equal to the RHS.
 \[LHS = RHS\]

Note: If the question involves the word prove just we have to show that the term which is in the LHS is equal to the term RHS with the help of or by using the algebraic identities. Since the question involves the algebraic term we must know about the identities of the algebraic terms.