Question

if a+b+c=9 and a²+b²+c²=35 find the value of x³+y³+z³-3xyz​

Answer 1

Given

• a+b+c=9
• a²+b²+c²=35

To Find

• a³+b³+c³-3abc

We know that,

${\sf →(a+b+c)²=a²+b²+c²+2ab+2bc+2ca}$

• on substituting the value

${\sf →(9)²=35+2ab+2bc+2ca}$

⁣${\sf →2ab+2bc+2ca= 81 -35}$

${\sf →2(ab+bc+ca)= 46}$

${\sf →ab+bc+ca= 23}$

Now,

${\sf →a³+b³+c³-3abc =(a+b+c)(a²+b²+c²-ab-bc-ca)}$

${\sf →a³+b³+c³-3abc =(a+b+c)(a²+b²+c²-(ab+bc+ca))}$

• on substituting the value

${\sf →a³+b³+c³-3abc =(9)(35-(23))}$

${\sf →a³+b³+c³-3abc =9×12}$

${\sf →a³+b³+c³-3abc = 108}$

$[/tex]</p><h3>Hence,</h3><p></p><ul><li>If a+b+c=9 and a²+b²+c²=35 then the value of a³+b³+c³-3abc will be <strong>108</strong>.</li></ul><p></p><p></p><p></p><p> </p><p> </p><p></p><ul><li>[tex]\begin{gathered}\begin{gathered}\blue{\begin{gathered}\tiny\begin{gathered}\small{\small{\small{\small{\small{\small{\small{\small{\small{\small{\begin{gathered}\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\red{ \bigstar} \: \underline{\bf{\orange{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\\\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \\\dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab \\\\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}}}}}}}}}}\end{gathered}\end{gathered}}\end{gathered}\end{gathered}$