Answer

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Hint: we are going to use the basic algebraic formula to solve the given question.

$$\because {a^3} + {b^3} + {c^3} - 3abc$$

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

$ = {a^2} + {b^2} + {c^2}$

$ = ab + bc + ca$ (given)

${a^3} + {b^3} + {c^3} - 3abc = 0 \times (a + b + c)$

$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0$

$\therefore {a^3} + {b^3} + {c^3} = 3abc$

Note: In algebra, we use alphabets like (x, y, z, a, b, c...) to substitute numbers in the equation to get the desired solution. Numbers are definite and their values are known. While alphabets are used to represent unknown numbers.

$$\because {a^3} + {b^3} + {c^3} - 3abc$$

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

$ = {a^2} + {b^2} + {c^2}$

$ = ab + bc + ca$ (given)

${a^3} + {b^3} + {c^3} - 3abc = 0 \times (a + b + c)$

$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0$

$\therefore {a^3} + {b^3} + {c^3} = 3abc$

Note: In algebra, we use alphabets like (x, y, z, a, b, c...) to substitute numbers in the equation to get the desired solution. Numbers are definite and their values are known. While alphabets are used to represent unknown numbers.

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