Answer
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Hint: The given problem is related to algebraic identities. $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)$ is of the form \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\] and here \[a+b+c=0\]. So, the identity \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\], when \[a+b+c=0\], is applicable here.
Complete step-by-step answer:
Let’s consider three numbers $a,b$ and $c$, such that \[a+b+c=0\]. We have to find the value of \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\].
We know \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)\]. We can write \[{{\left( a+b+c \right)}^{3}}\] as \[{{\left( a+b+c \right)}^{2}}\left( a+b+c \right)\].
So, ${{\left( a+b+c \right)}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \right)$ .
On multiplying the terms in the right-hand side of the equation, we get:
${{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+{{a}^{2}}\left( b+c \right)+{{b}^{2}}\left( c+a \right)+{{c}^{2}}\left( b+a \right)+2\left( ab+bc+ca \right)\left( a+b+c \right)$
$\Rightarrow {{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3{{a}^{2}}(b+c)+3{{b}^{2}}(a+c)+3{{c}^{2}}(a+b)+6abc$
Shifting \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\] to left-hand side of the equation and rest of the terms to right-hand side of the equation, we get:
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b+c \right)}^{3}}-3{{a}^{2}}(b+c)-3{{b}^{2}}(a+c)-3{{c}^{2}}(a+b)-6abc\]
Now, we will subtract $3abc$ from both sides of the equation. So, we get:
$\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{\left( a+b+c \right)}^{3}}-\left[ 3{{a}^{2}}(b+c)+3{{b}^{2}}(a+c)+3{{c}^{2}}(a+b)+9abc \right]$
$\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{\left( a+b+c \right)}^{3}}-\left[ 3{{a}^{2}}b+3{{a}^{2}}c+3abc+3{{b}^{2}}a+3{{b}^{2}}c+3abc+3{{c}^{2}}a+3{{c}^{2}}b+3abc \right]$
Now, on rearranging the terms, we get:
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{\left( a+b+c \right)}^{3}}-\left[ 3{{a}^{2}}b+3{{b}^{2}}a+3abc+3{{b}^{2}}c+3{{c}^{2}}b+3abc+3{{c}^{2}}a+3{{a}^{2}}c+3abc \right]$
Now, taking $3ab,3bc$ and $3ca$ common, and substituting ${{\left( a+b+c \right)}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \right)$ , we get:
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \right)-\left[ 3ab\left( a+b+c \right)+3bc\left( a+b+c \right)+3ca\left( a+b+c \right) \right]\]
Taking $\left( a+b+c \right)$ common, we get
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)......(i)\]
Now, when $(a+b+c)=0$ , equation $(i)$ becomes \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\].
Now, we will consider the given problem. We are asked to find the value of $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)$ without actually calculating the cubes. Comparing ${{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}}$ with \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\], we get $a=-12,b=7$ and $c=5$. On calculating the value of $(a+b+c)$ , we can see $a+b+c=-12+7+5=0$. As $a+b+c=0$ , so, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\].
Or, $-\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}} \right)=-3abc$.
So, $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)=-3\times \left( -12 \right)\times 7\times 5=1260$.
Hence, the value of $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)$ is equal to $1260$.
Note: Most of the students remember the formula \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\], but do not remember the condition for the formula to be true, i.e. $(a+b+c)=0$ which can be very dangerous. So, instead of remembering \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\], it is better to remember the complete formula \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)\]. It is true for any condition and thus, the student will be on a safe side and can use this formula in any condition.
Complete step-by-step answer:
Let’s consider three numbers $a,b$ and $c$, such that \[a+b+c=0\]. We have to find the value of \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\].
We know \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)\]. We can write \[{{\left( a+b+c \right)}^{3}}\] as \[{{\left( a+b+c \right)}^{2}}\left( a+b+c \right)\].
So, ${{\left( a+b+c \right)}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \right)$ .
On multiplying the terms in the right-hand side of the equation, we get:
${{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+{{a}^{2}}\left( b+c \right)+{{b}^{2}}\left( c+a \right)+{{c}^{2}}\left( b+a \right)+2\left( ab+bc+ca \right)\left( a+b+c \right)$
$\Rightarrow {{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3{{a}^{2}}(b+c)+3{{b}^{2}}(a+c)+3{{c}^{2}}(a+b)+6abc$
Shifting \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\] to left-hand side of the equation and rest of the terms to right-hand side of the equation, we get:
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b+c \right)}^{3}}-3{{a}^{2}}(b+c)-3{{b}^{2}}(a+c)-3{{c}^{2}}(a+b)-6abc\]
Now, we will subtract $3abc$ from both sides of the equation. So, we get:
$\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{\left( a+b+c \right)}^{3}}-\left[ 3{{a}^{2}}(b+c)+3{{b}^{2}}(a+c)+3{{c}^{2}}(a+b)+9abc \right]$
$\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{\left( a+b+c \right)}^{3}}-\left[ 3{{a}^{2}}b+3{{a}^{2}}c+3abc+3{{b}^{2}}a+3{{b}^{2}}c+3abc+3{{c}^{2}}a+3{{c}^{2}}b+3abc \right]$
Now, on rearranging the terms, we get:
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{\left( a+b+c \right)}^{3}}-\left[ 3{{a}^{2}}b+3{{b}^{2}}a+3abc+3{{b}^{2}}c+3{{c}^{2}}b+3abc+3{{c}^{2}}a+3{{a}^{2}}c+3abc \right]$
Now, taking $3ab,3bc$ and $3ca$ common, and substituting ${{\left( a+b+c \right)}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \right)$ , we get:
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \right)-\left[ 3ab\left( a+b+c \right)+3bc\left( a+b+c \right)+3ca\left( a+b+c \right) \right]\]
Taking $\left( a+b+c \right)$ common, we get
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)......(i)\]
Now, when $(a+b+c)=0$ , equation $(i)$ becomes \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\].
Now, we will consider the given problem. We are asked to find the value of $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)$ without actually calculating the cubes. Comparing ${{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}}$ with \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\], we get $a=-12,b=7$ and $c=5$. On calculating the value of $(a+b+c)$ , we can see $a+b+c=-12+7+5=0$. As $a+b+c=0$ , so, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\].
Or, $-\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}} \right)=-3abc$.
So, $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)=-3\times \left( -12 \right)\times 7\times 5=1260$.
Hence, the value of $-\left( {{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}} \right)$ is equal to $1260$.
Note: Most of the students remember the formula \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\], but do not remember the condition for the formula to be true, i.e. $(a+b+c)=0$ which can be very dangerous. So, instead of remembering \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\], it is better to remember the complete formula \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)\]. It is true for any condition and thus, the student will be on a safe side and can use this formula in any condition.
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