Solve b^4+b^3+343b+343

$$b ^ { 4 } + b ^ { 3 } + 343 b + 343$$

Factor

$$\left(b+1\right)\left(b+7\right)\left(b^{2}-7b+49\right)$$

Solution Steps

Do the grouping b^{4}+b^{3}+343b+343=\left(b^{4}+b^{3}\right)+\left(343b+343\right), and factor out b^{3} in the first and 343 in the second group.

$$b^{3}\left(b+1\right)+343\left(b+1\right)$$

Factor out common term b+1 by using distributive property.

$$\left(b+1\right)\left(b^{3}+343\right)$$

Consider b^{3}+343. Rewrite b^{3}+343 as b^{3}+7^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).

$$\left(b+7\right)\left(b^{2}-7b+49\right)$$

Rewrite the complete factored expression. Polynomial b^{2}-7b+49 is not factored since it does not have any rational roots.

$$\left(b+1\right)\left(b+7\right)\left(b^{2}-7b+49\right)$$

Evaluate

$$b^{4}+b^{3}+343b+343$$