
Factorize ${\text{ 64}}{{\text{m}}^3} - 343{n^3}$
A. $\left( {4m + 7n} \right)\left( {16{m^2} - 28mn + 49{n^2}} \right)$
B. $\left( {4m - 7n} \right)\left( {16{m^2} + 28mn + 49{n^2}} \right)$
C. $\left( {4m + 7n} \right)\left( {16{m^2} + 28mn + 49{n^2}} \right)$
D. $\left( {4m - 7n} \right)\left( {16{m^2} - 28mn + 49{n^2}} \right)$
Answer
483.9k+ views
Hint: In order to solve this question the key observation is it is of the form of the identity given by,
${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$. After comparing it with this standard identity the given question can be factorized easily and hence be solved.
Complete step-by-step answer:
The question given is to factorize ${\text{64}}{{\text{m}}^3} - 343{n^3}$
$ 64 = {4^3}$
And, $343 = {7^3}$
${\text{64}}{{\text{m}}^3} - 343{n^3}$ can be written as,
${\text{64}}{{\text{m}}^3} - 343{n^3} = {4^3}{m^3} - {7^3}{n^3}$
Or,
${\text{64}}{{\text{m}}^3} - 343{n^3} = {\left( {4m} \right)^3} - {\left( {7n} \right)^3}$
$ {\left( {4m} \right)^3} - {\left( {7n} \right)^3}{\text{ is of the form }}{a^3} - {b^3}$
Using the identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
$\Rightarrow $ ${\left( {4m} \right)^3} - {\left( {7n} \right)^3}$ can be written as,
${\left( {4m} \right)^3} - {\left( {7n} \right)^3} = \left( {4m - 7n} \right)\left( {{{\left( {4m} \right)}^2} + {{\left( {7n} \right)}^2} + 4m \times 7n} \right)$
On simplifying further,
${\left( {4m} \right)^3} - {\left( {7n} \right)^3} = \left( {4m - 7n} \right)\left( {16{m^2} + 49{n^2} + 28mn} \right)$
$\Rightarrow $ Option (B) is correct.
Note: The factored form can be checked for its correctness by simply expanding and multiplying, which would result in the answer matching the original and simplified form. If the question asks to faction the equation of the form ${a^3} + {b^3}$ then it can be done using the identity given by ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$.The identities relation, related to solve the factorization problems must be remembered to use it directly wherever necessary. The units can be checked or the coefficients can be checked whether to apply the identities of cubic functions or something else.
${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$. After comparing it with this standard identity the given question can be factorized easily and hence be solved.
Complete step-by-step answer:
The question given is to factorize ${\text{64}}{{\text{m}}^3} - 343{n^3}$
$ 64 = {4^3}$
And, $343 = {7^3}$
${\text{64}}{{\text{m}}^3} - 343{n^3}$ can be written as,
${\text{64}}{{\text{m}}^3} - 343{n^3} = {4^3}{m^3} - {7^3}{n^3}$
Or,
${\text{64}}{{\text{m}}^3} - 343{n^3} = {\left( {4m} \right)^3} - {\left( {7n} \right)^3}$
$ {\left( {4m} \right)^3} - {\left( {7n} \right)^3}{\text{ is of the form }}{a^3} - {b^3}$
Using the identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
$\Rightarrow $ ${\left( {4m} \right)^3} - {\left( {7n} \right)^3}$ can be written as,
${\left( {4m} \right)^3} - {\left( {7n} \right)^3} = \left( {4m - 7n} \right)\left( {{{\left( {4m} \right)}^2} + {{\left( {7n} \right)}^2} + 4m \times 7n} \right)$
On simplifying further,
${\left( {4m} \right)^3} - {\left( {7n} \right)^3} = \left( {4m - 7n} \right)\left( {16{m^2} + 49{n^2} + 28mn} \right)$
$\Rightarrow $ Option (B) is correct.
Note: The factored form can be checked for its correctness by simply expanding and multiplying, which would result in the answer matching the original and simplified form. If the question asks to faction the equation of the form ${a^3} + {b^3}$ then it can be done using the identity given by ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$.The identities relation, related to solve the factorization problems must be remembered to use it directly wherever necessary. The units can be checked or the coefficients can be checked whether to apply the identities of cubic functions or something else.
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