Question

factorise
1) 81n⁴-p⁴. Answer will be (3n+p) (3n-p) (9n²+p²)​

Answer 1

Answer:

Yes your answer is correct.

$\Large\boxed{\longrightarrow\mathsf{(9n^2+p^2)(3n+p)(3n-p)}}$

Step-by-step explanation:

GIVEN:

81n⁴-p⁴

TO FIND:

The factorise of 81n⁴-p⁴.

TO SOLVE:

With exponent rules and difference of two squares.

SOLUTIONS:

First, rewrite the equation problem.

$\displaystyle \mathsf{9^2=9\times9=81}$

$\displaystyle \mathsf{9^2n^4-p^4}$

Use exponent rule.

$\Large\boxed{\mathsf{EXPONENT \quad  RULES}}$

$\displaystyle \mathsf{A^B^C=(A^B)^C}$

$\displaystyle \mathsf{(P^2)^2=P^4}$

$\displaystyle \mathsf{9^2n^4-\left(p^2\right)^2}}}$

$\displaystyle \mathsf{(n^2)^2=n^4}$

$\displaystyle \mathsf{9^2\left(n^2\right)^2-\left(p^2\right)^2}}}}}$

Exponent rules.

$\displaystyle \mathsf{A^MB^M=(AB)^M}$

$\displaystyle \mathsf{9^2\left(n^2\right)^2=\left(9n^2\right)^2}$

$\displaystyle \mathsf{\left(9n^2\right)^2-\left(p^2\right)^2}}}$

Use difference from between the two squares.

$\Large\boxed{\mathsf{TWO \quad SQUARES \quad FORMULA}}$

$\displaystyle \mathsf{X^2-Y^2=(X+Y)(X-Y)}$

$\displaystyle \mathsf{\left(9n^2\right)^2-\left(p^2\right)^2=\left(9n^2+p^2\right)\left(9n^2-p^2\right)}}$

$\displaystyle \mathsf{\left(9n^2+p^2\right)\left(9n^2-p^2\right)}$

Factors of 9n²-p².

Solve.

$\displaystyle \mathsf{9n^2-p^2=\boxed{\mathsf{(3n+p)(3n-p)}}}}$

$\Large\boxed{\longrightarrow\mathsf{(9n^2+p^2)(3n+p)(3n-p)}}}}$

Hence, the final answer is (9n²+p²)(3n+p)(3n-p).