
Verify the following expression (i) $({x^3} + {y^3}) = (x + y)({x^2} - xy + {y^2})$
Answer
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Hint: According to given in the question to verify (i) $({x^3} + {y^3}) = (x + y)({x^2} - xy + {y^2})$ we have to obtain the expression in R.H.S by solving L.H.S.
So, to solve the L.H.S we have to use the formula as given below:
Formula used: $ \Rightarrow {(a - b)^3} = {a^3} + {b^3} + 3ab(a + b)..................(1)$
Hence, after applying the formula mentioned above we have to solve the expression obtained now, to solve the expression obtained we have to use the formula as mentioned below:
$ \Rightarrow {(a - b)^2} = {a^2} + {b^2} + 2ab..............(2)$
Hence, after applying the formula mentioned above we obtained R.H.S
Complete step-by-step solution:
Step 1: To verify first of all we have to simplify L.H.S to obtain R.H.S with the help of formula 1 as mentioned in the solution hint.
$
\Rightarrow {(x - y)^3} = {x^3} + {y^3} + 3xy(x + y) \\
\Rightarrow {x^3} + {y^3} = {(x - y)^3} - 3xy(x + y)
$
Now, taking (x-y) as common from the expression obtained just above,
$ = (x - y)\left[ {{{(x - y)}^2} - 3xy} \right]...............(2)$
Step 2: Now, to solve the expression (2) obtained in step to we have to apply the formula (2) as mentioned in the solution hint.
$
= (x - y)\left[ {{{(x - y)}^2} - 3xy} \right] \\
= (x - y)\left[ {{x^2} + {y^2} - 2xy - 3xy} \right] \\
= (x - y)({x^2} + {y^2} - xy)
$
Which is equal to R.H.S
Hence, with the help of formula (1) we have verified $({x^3} + {y^3}) = (x + y)({x^2} - xy + {y^2})$
Note: We can also solve this problem by considering RHS first and showing that the RHS is equal to LHS. In this method we just have to expand the brackets and consider all the like terms together. To verify these types of expressions it is necessary to break them into parts or solve them into parts so for each part we can apply the necessary formula to solve if required.
So, to solve the L.H.S we have to use the formula as given below:
Formula used: $ \Rightarrow {(a - b)^3} = {a^3} + {b^3} + 3ab(a + b)..................(1)$
Hence, after applying the formula mentioned above we have to solve the expression obtained now, to solve the expression obtained we have to use the formula as mentioned below:
$ \Rightarrow {(a - b)^2} = {a^2} + {b^2} + 2ab..............(2)$
Hence, after applying the formula mentioned above we obtained R.H.S
Complete step-by-step solution:
Step 1: To verify first of all we have to simplify L.H.S to obtain R.H.S with the help of formula 1 as mentioned in the solution hint.
$
\Rightarrow {(x - y)^3} = {x^3} + {y^3} + 3xy(x + y) \\
\Rightarrow {x^3} + {y^3} = {(x - y)^3} - 3xy(x + y)
$
Now, taking (x-y) as common from the expression obtained just above,
$ = (x - y)\left[ {{{(x - y)}^2} - 3xy} \right]...............(2)$
Step 2: Now, to solve the expression (2) obtained in step to we have to apply the formula (2) as mentioned in the solution hint.
$
= (x - y)\left[ {{{(x - y)}^2} - 3xy} \right] \\
= (x - y)\left[ {{x^2} + {y^2} - 2xy - 3xy} \right] \\
= (x - y)({x^2} + {y^2} - xy)
$
Which is equal to R.H.S
Hence, with the help of formula (1) we have verified $({x^3} + {y^3}) = (x + y)({x^2} - xy + {y^2})$
Note: We can also solve this problem by considering RHS first and showing that the RHS is equal to LHS. In this method we just have to expand the brackets and consider all the like terms together. To verify these types of expressions it is necessary to break them into parts or solve them into parts so for each part we can apply the necessary formula to solve if required.
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