Question

Prove

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = ( {a}^{2} + {b }^{2} + {c}^{2} -ab - bc - ca) \times ( a+ b + c )$
100 points
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Answer 1

here ,

RHS =

$({a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)(a + b + c) \\ \\ = {a}^{2} (a + b + c) + {b}^{2} (a + b + c) + {c}^{2} (a + b + c) - ab(a + b + c) - bc(a + b + c) - ca(a + b + c) \\ \\ now \: simplify \: above \: eqution \: \\ \\ = ({a}^{3} + {a}^{2} b + {a}^{2} c) + ( {b}^{2} a + {b}^{3} + {b}^{2} c) + ( {c}^{2} a + {c}^{2} b + {c}^{3} ) + ( - {a}^{2} b - {b}^{2} a - abc) + ( - abc - {b}^{2} c - {c}^{2} b) + ( - {a}^{2} c - abc - {c}^{2} a) \\ \\ now \: remove \: brakets \\ \\ = {a}^{3} + {a}^{2} b + {a}^{2} c + {b}^{2} a + {b}^{3} + {b}^{2} c + {c}^{2} a + {c}^{2} b + {c}^{3} - {a}^{2} b - {b}^{2} a - abc - abc - {b}^{2} c - {c}^{2} b - {a}^{2} c - abc - {c}^{2} a \\ \\ = {a}^{3} + {b}^{3} + {c}^{3} - abc - abc - abc \\ \\ = {a}^{3} + {b}^{3} + {c}^{3} - 3abc \\ \\ because \: others \: terms \: are \: oposite \: sign \: to \: each \: other \: so \: it \: is \: cancelled \: out \\ \\ so \: we \: can \: show \: that \\ \\ = {a}^{3} + {b}^{3} + {c}^{3} - 3abc$

so RHS = LHS

terms are so long so you swipe your screen left and right...

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Answer 2

Simplify:

Step-by-step explanation:

$\ Given \ value:\\\\a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-bc-ca)\times (a+b+c)\\\\\ Solution:\\\\\ solve \ R.H.S \ part:\\\\(a^2+b^2+c^2-ab-bc-ca)\times (a+b+c)\\\\\rightarrow a^3+a^2b+a^2c+ab^2+b^3+b^2c+ac^2+bc^2+c^3-a^2b-ab^2-abc-abc-b^2c-b^2c^2-ca^2-abc-c^2a\\\\\rightarrow a^3+a^2b+a^2c+ab^2+b^3+b^2c+ac^2+bc^2+c^3-a^2b-ab^2-3abc-b^2c-b^2c^2-ca^2-c^2a\\\\ \ by \ add \ and \ subtract \ we \ get \\\\\rightarrow a^3+b^3+c^3-3abc$

Learn more:

• Simplify: https://brainly.in/question/8451502