if a = 2\sqrt{2}2 + 1, b = 3\sqrt{3}3 - 2\sqrt{2}2 and c = 6\sqrt{6}6 + 2 2\sqrt{2}2 then the value of a2a^{2}a2 + b2b^{2}b2 - c is
asked by guest on Apr 04, 2025 at 11:01 pm
MathBot Answer:
a2+b2−c=−22−6+(−2+3)2+(1+2)2{a}^{2} + {b}^{2} - c = - 2 \sqrt{2} - \sqrt{6} + \left(- \sqrt{2} + \sqrt{3}\right)^{2} + \left(1 + \sqrt{2}\right)^{2}a2+b2−c=−22−6+(−2+3)2+(1+2)2
a2+b2−c=(2+1)2+(3−2)2−6+22=(3−2)2+(2+1)2−6+22=(2+1)2+(3−2)2−6−22=−22−6+(−2+3)2+(1+2)2\begin{aligned}a^{2} + b^{2} - c&=\left(\sqrt{2} + 1\right)^{2} + \left(\sqrt{3} - \sqrt{2}\right)^{2} - \sqrt{6} + 2 \sqrt{2}\\&=\left(\sqrt{3} - \sqrt{2}\right)^{2} + \left(\sqrt{2} + 1\right)^{2} - \sqrt{6} + 2 \sqrt{2}\\&=\left(\sqrt{2} + 1\right)^{2} + \left(\sqrt{3} - \sqrt{2}\right)^{2} - \sqrt{6} - 2 \sqrt{2}\\&=- 2 \sqrt{2} - \sqrt{6} + \left(- \sqrt{2} + \sqrt{3}\right)^{2} + \left(1 + \sqrt{2}\right)^{2}\end{aligned}a2+b2−c=(2+1)2+(3−2)2−6+22=(3−2)2+(2+1)2−6+22=(2+1)2+(3−2)2−6−22=−22−6+(−2+3)2+(1+2)2
