$48^{2} + 53^{2} + 62^{2} =$ $48^2+53^2+62^2=$ ?

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Answer 1 · 2015-12-25T00:16:35

$8957$ $8957$

Use ${(a \pm b)}^{2} = a^{2} \pm 2 a b + b^{2}$ $(a+-b)^2 = a^2+-2ab+b^2$

$48^{2} = {(50 - 2)}^{2} = 50^{2} - 2 (50) (2) + 2^{2}$ $48^2 = (50-2)^2 = 50^2-2(50)(2)+2^2$

$= 2500 - 200 + 4 = 2304$ $= 2500-200+4 = color(blue)(2304)$

$53^{2} = {(50 + 3)}^{2} = 50^{2} + 2 (50) (3) + 3^{2}$ $53^2 = (50+3)^2 = 50^2+2(50)(3)+3^2$

$= 2500 + 300 + 9 = 2809$ $=2500+300+9 = color(blue)(2809)$

$62^{2} = {(50 + 12)}^{2} = 50^{2} + 2 (50) (12) + 12^{2}$ $62^2 = (50+12)^2 = 50^2+2(50)(12)+12^2$

$= 2500 + 1200 + 144 = 3844$ $=2500+1200+144 = color(blue)(3844)$

So:

$48^{2} + 53^{2} + 62^{2} = 2304 + 2809 + 3844 = 8957$ $48^2+53^2+62^2 = 2304+2809+3844 = 8957$

482+532+622=48^2+53^2+62^2=?