Question

add a³ +b³+c³- abc, b³-4c² +6 abc+3a² and+ c³+5abc​

Answer 1

$\LARGE\textsf{\underline\textcolor{aqua}{↭ SoLuTioN :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{( a³ + b³ + c³ - abc ) + ( b³ - 4c² + 6abc + 3a³) + ( c³ + 5abc )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ + b³ + c³ - abc + b³ - 4c² + 6abc + 3a³+ c³ + 5abc }$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ + 3a³ + b³ + b³ + c³ - 4c³ + c³ - abc + 6abc + 5abc}$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{4a³ + 2b³ - 2c³ + 10abc}}$

$\large\textsf{ }$

$\LARGE\textsf{\underline\textcolor{aqua}{↭ MoRe fOrMuLaS :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{( a + b )² = a² + b² + 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{( a - b )² = a² + b² - 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{a² - b² = ( a + b ) ( a - b )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a² + b² = ( a + b )² - 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ + b³ = ( a + b ) ( a² - 2ab + b² )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ - b³ = ( a - b ) ( a² + ab + b² )}$

$\qquad\tt{:}\longrightarrow\large\textsf{( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ac }$

$\large\textsf{ }$

$\large\textsf\textcolor{purple}{ \; \; \; \; \; \; \; \; \; \; \; \; ◈ ━━━━━━━ ✪ ━━━━━━━ ◈}$

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Answer 2

There is a formula

a^3+b^3+c^3–3abc

=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

If a^2+b^2+c^2-ab-bc-ca= 0 then

a^3+b^3+c^3–3abc= 0 => a^3+b^3+c^3 = 3abc