How do you factor \[125{x^6} - 8{y^6}\]?
Answer
Verified
440.1k+ 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.
Recently Updated Pages
What is fractional part in 0707 15 20 class 10 maths CBSE
Find the square root of 1225 using a long division class 10 maths CBSE
Who were considered as inferior and undesirable by class 10 social science CBSE
A river 3m deep and 40m wide is flowing at the rate class 10 maths CBSE
In what ratio is the line segment joining the poin-class-10-maths-CBSE
A rectangular sheet of paper is rolled along its length class 10 maths CBSE
Trending doubts
Assertion The planet Neptune appears blue in colour class 10 social science CBSE
The term disaster is derived from language AGreek BArabic class 10 social science CBSE
Imagine that you have the opportunity to interview class 10 english CBSE
Find the area of the minor segment of a circle of radius class 10 maths CBSE
Differentiate between natural and artificial ecosy class 10 biology CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE