How do you factor \[125{x^6} - 8{y^6}\]?
Answer
620.7k+ views
Hint: Here in this question, we have to find the factors of the given equation. If you see the equation it is in the form of \[{a^3} + {b^3}\]. We have a standard formula on this algebraic equation and it is given by \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], hence by substituting the value of a and b we find the factors.
Complete step-by-step solution:
The equation is an algebraic equation or expression, where algebraic expression is a combination of variables and constants.
Now consider the given equation \[125{x^6} - 8{y^6}\], let we write in the exponential form. The number \[125{x^6}\] can be written as \[5{x^2} \times 5{x^2} \times 5{x^2}\] and the \[8{y^6}\]can be written as \[2{y^2} \times 2{y^2} \times 2{y^2}\], in the exponential form it is \[{\left( {2{y^2}} \right)^3}\]. The number \[125{x^6}\] is written as \[5{x^2} \times 5{x^2} \times 5{x^2}\] and in exponential form is \[{(5{x^2})^3}\].
Therefore, the given equation is written as \[{\left( {5{x^2}} \right)^3} - {(2{y^2})^3}\], the equation is in the form of \[{a^3} - {b^3}\].The \[{a^3} - {b^3}\]have a standard formula on this algebraic equation and it is given by \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\], here the value of a is \[5{x^2}\] and the value of b is \[2{y^2}\] . By substituting these values in the formula, we have
\[125{x^6} - 8{y^6} = {\left( {5{x^2}} \right)^3} - {\left( {2{y^2}} \right)^3} = \left( {5{x^2} - 2{y^2}} \right)\left( {{{\left( {5{x^2}} \right)}^2} + (5{x^2})(2{y^2}) + {{\left( {2{y^2}} \right)}^2}} \right)\]
On simplifying we have
\[ \Rightarrow 125{x^6} - 8{y^6} = (5{x^2} - 2{y^2})(25{x^4} + 10{x^2}{y^2} + 4{y^4})\]
The second term can’t be simplified further. Since it contains the two terms which are unknown. Therefore we keep the second term as it is. Therefore, the factors of \[125{x^6} - 8{y^6}\] is \[(5{x^2} - 2{y^2})(25{x^4} + 10{x^2}{y^2} + 4{y^4})\].
Note: To find the factors for algebraic equations or expressions, it depends on the degree of the equation. If the equation contains a square then we have two factors. If the equation contains a cube then we have three factors. Here this equation also contains 3 factors, the two factors are imaginary.
Complete step-by-step solution:
The equation is an algebraic equation or expression, where algebraic expression is a combination of variables and constants.
Now consider the given equation \[125{x^6} - 8{y^6}\], let we write in the exponential form. The number \[125{x^6}\] can be written as \[5{x^2} \times 5{x^2} \times 5{x^2}\] and the \[8{y^6}\]can be written as \[2{y^2} \times 2{y^2} \times 2{y^2}\], in the exponential form it is \[{\left( {2{y^2}} \right)^3}\]. The number \[125{x^6}\] is written as \[5{x^2} \times 5{x^2} \times 5{x^2}\] and in exponential form is \[{(5{x^2})^3}\].
Therefore, the given equation is written as \[{\left( {5{x^2}} \right)^3} - {(2{y^2})^3}\], the equation is in the form of \[{a^3} - {b^3}\].The \[{a^3} - {b^3}\]have a standard formula on this algebraic equation and it is given by \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\], here the value of a is \[5{x^2}\] and the value of b is \[2{y^2}\] . By substituting these values in the formula, we have
\[125{x^6} - 8{y^6} = {\left( {5{x^2}} \right)^3} - {\left( {2{y^2}} \right)^3} = \left( {5{x^2} - 2{y^2}} \right)\left( {{{\left( {5{x^2}} \right)}^2} + (5{x^2})(2{y^2}) + {{\left( {2{y^2}} \right)}^2}} \right)\]
On simplifying we have
\[ \Rightarrow 125{x^6} - 8{y^6} = (5{x^2} - 2{y^2})(25{x^4} + 10{x^2}{y^2} + 4{y^4})\]
The second term can’t be simplified further. Since it contains the two terms which are unknown. Therefore we keep the second term as it is. Therefore, the factors of \[125{x^6} - 8{y^6}\] is \[(5{x^2} - 2{y^2})(25{x^4} + 10{x^2}{y^2} + 4{y^4})\].
Note: To find the factors for algebraic equations or expressions, it depends on the degree of the equation. If the equation contains a square then we have two factors. If the equation contains a cube then we have three factors. Here this equation also contains 3 factors, the two factors are imaginary.
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