Evaluate the product of $b^{4} (b^{6} + b^{8})$

Open in App

Solution

The given polynomials are $b^{4}$ and $b^{6} + b^{8}$ we multiply them as follows:

$(b^{4}) (b^{6} + b^{8}) = (b^{4} \times b^{6}) + (b^{4} \times b^{8}) = b^{10} + b^{12}$

Hence, $(b^{4}) (b^{6} + b^{8}) = b^{10} + b^{12}$ .

Suggest Corrections

Similar questions

Q. ax+b8.

a^{8} - 2 a^{4} b^{4} + b^{8} =

$I f \frac{s i n^{4} θ}{a} + \frac{c o s^{4} θ}{b} = \frac{1}{a + b^{'}}, p r o v e t h a t \frac{s i n^{8} θ}{a^{8}} + \frac{c o s^{8} θ}{b^{8}} = \frac{1}{a + b^{8}}$

Q. Factorise the following.

256 a^{8} - b^{8}

Q. Solve:

12 a^{8} b^{8} \div (- 6 a^{6} b^{4})

Join BYJU'S Learning Program

Evaluate the product of b4(b6+b8)

The given polynomials are b4 and b6+b8 we multiply them as follows:(b4)(b6+b8)=(b4×b6)+(b4×b8)=b10+b12Hence, (b4)(b6+b8)=b10+b12.

Evaluate the product of $b^{4} (b^{6} + b^{8})$

The given polynomials are $b^{4}$ and $b^{6} + b^{8}$ we multiply them as follows:

$(b^{4}) (b^{6} + b^{8}) = (b^{4} \times b^{6}) + (b^{4} \times b^{8}) = b^{10} + b^{12}$

Hence, $(b^{4}) (b^{6} + b^{8}) = b^{10} + b^{12}$ .