Question

Find the value of x3 + y3 + z3

– 3xyz, if x + y + z = 12 and x

2 + y2 + z2 =70
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Answer 1

Answer:

396

Step-by-step explanation:

for detailed answer see the image tagged with it

Answer 2

$\red{ \bigstar} \underline{\underline{ \large{ \sf {given}}}}$

$\longrightarrow \tt{x + y +z = 12 }$

$\longrightarrow \tt{ {x}^{2} + {y}^{2} + {z}^{2} = 70}$

$\red{ \bigstar}\underline{ \underline{ \large{ \sf{to \: find}}}}$

$\longrightarrow \tt{ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz }$

$\red { \bigstar}\underline{ \underline{ \large{ \sf{algebraic \: identities \: used}}}}$

$\star \tt{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac) }$

$\star \: \tt{ {a}^{3} + {b}^{3} + {c}^{3} - 3abc= (a+ b+ c)( {a}^{2} + {b}^{2} + {c}^{2}) - (ab+ bc + ac)) }$

$\red{ \bigstar}\underline{ \underline{ \large{ \sf{solution}}}}$

$\blue{ \tt{using \: algebraic\: identity}}$

$\tt{ {(x + y + z)}^{2} = ({x}^{2} + {y}^{2} + {z}^{2}) + 2(xy + yz + xz) }$

$\blue{ \tt{putting \: known \: values}}$

$\tt{ {(12)}^{2} = (70) + 2(xy + yz + xz) }$

$\tt{144 - 70 = 2(xy + yz + xz)}$

$\pink{ \tt{xy + yz + xz = 37}}$

Now,

again

$\blue{ \tt{using \: algebraic \: identity \: }}$

$\tt{ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( ({x}^{2} + {y}^{2} + {z}^{2} ) - (xy + yz + xz))}$

$\blue{ \tt{putting \: known \: values}}$

$\tt{ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (12)(70 - 37) }$

$\boxed{ \tt{ \pink{ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 396}}}$

Answer.