Answer:
14
Step-by-step explanation:
Evaluate
[tex](1+\sqrt{2})^3 + (1-\sqrt{2})^3[/tex]
As we have the Cube binomials:
[tex]\boxed{(a+b)^3=a^3+3a^2b+3ab^2+b^3}[/tex]
[tex]\boxed{(a-b)^3=a^3-3a^2b+3ab^2-b^3}[/tex]
[tex](1+\sqrt{2})^3 =1^3 + 3\cdot 1^2 \cdot\sqrt{2} + 3 \cdot 1 \cdot (\sqrt{2}) ^2 + (\sqrt{2}) ^3 = \boxed{7+5\sqrt{2}}[/tex]
[tex](1-\sqrt{2})^3 =1^3 - 3\cdot 1^2 \cdot\sqrt{2} + 3 \cdot 1 \cdot (\sqrt{2}) ^2 - (\sqrt{2}) ^3 = \boxed{7-5\sqrt{2}}[/tex]
[tex]\therefore (1+\sqrt{2})^3 + (1-\sqrt{2})^3 =7+5\sqrt{2} + 7-5\sqrt{2} = \boxed{14}[/tex]
