Question

please help me please

3 If a + b + c = 5 and ab + bc + ca = 10, prove that a3 + b3 + c3 - 3abc = -25.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a + b + c = 5 - - - - (1)$

and

$\rm :\longmapsto\:ab + bc + ca = 10 - - - - (2)$

Now, Consider

$\rm :\longmapsto\:a + b + c = 5$

On squaring both sides, we get

$\rm :\longmapsto\: {(a + b + c)}^{2} = {5}^{2}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 25$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 25$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(10) = 25 \: \: \: \{using \: equation \: (2) \}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 20 = 25$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 25 - 20$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 5 - - - - (3)$

Now, Consider

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + {c}^{3} - 3abc$

$\: \rm \: = \: (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

$\: \rm \: = \: (a + b + c)[{a}^{2} + {b}^{2} + {c}^{2} - (ab + bc + ca)]$

On substituting the values from equation (1), (2) and (3), we get

$\rm \:  =  \: (5)(5 - 10)$

$\rm \:  =  \: (5)( - 5)$

$\rm \:  =  \: - 25$

Hence,

$\purple{\rm\implies \:\boxed{\tt{ \: \: \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc = - 25 \: \: \: }}}$

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MORE IDENTITIES TO KNOW

$\green{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\green{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\blue{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\blue{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\gray{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

$\gray{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a + b + c = 5 - - - - (1)$

and

$\rm :\longmapsto\:ab + bc + ca = 10 - - - - (2)$

Now, Consider

$\rm :\longmapsto\:a + b + c = 5$

On squaring both sides, we get

$\rm :\longmapsto\: {(a + b + c)}^{2} = {5}^{2}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 25$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 25$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(10) = 25 \: \: \: \{using \: equation \: (2) \}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 20 = 25$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 25 - 20$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 5 - - - - (3)$

Now, Consider

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + {c}^{3} - 3abc$

$\: \rm \: = \: (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

$\: \rm \: = \: (a + b + c)[{a}^{2} + {b}^{2} + {c}^{2} - (ab + bc + ca)]$

On substituting the values from equation (1), (2) and (3), we get

$\rm \:  =  \: (5)(5 - 10)$

$\rm \:  =  \: (5)( - 5)$

$\rm \:  =  \: - 25$

Hence,

$\purple{\rm\implies \:\boxed{\tt{ \: \: \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc = - 25 \: \: \: }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE IDENTITIES TO KNOW

$\green{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\green{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\blue{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\blue{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\gray{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

$\gray{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$