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Question Number 9548 by Joel575 last updated on 14/Dec/16

a − 7b + 8c = 4 8a + 4b − c = 7 a, b, c ∈ R 60(a^2 − b^2 + c^2 ) = ?

$${a}\:−\:\mathrm{7}{b}\:+\:\mathrm{8}{c}\:=\:\mathrm{4} \\ $$$$\mathrm{8}{a}\:+\:\mathrm{4}{b}\:−\:{c}\:=\:\mathrm{7} \\ $$$${a},\:{b},\:{c}\:\in\:\mathbb{R} \\ $$$$ \\ $$$$\mathrm{60}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right)\:=\:? \\ $$

Answered by ridwan balatif last updated on 15/Dec/16

a−7b+8c=4 a+8c=4+7b (a+8c)^2 =(4+7b)^2 a^2 +16ac+64c^2 =16+56b+49b^2 ...(1) 8a+4b−c=7 8a−c=7−4b (8a−c)^2 =(7−4b)^2 64a^2 −16ac+c^2 =49−56b+16b^2 ...(2) (1)+(2) a^2 +16ac+64c^2 =16+56b+49b^2 64a^2 −16ac+c^2 =49−56b+16b^2 −−−−−−−−−−−−−−−−−−+ 65a^2 +65c^2 =65+65b^2 a^2 +c^2 =1+b^2 a^2 −b^2 +c^2 =1 60(a^2 −b^2 +c^2 )=60

$$\mathrm{a}−\mathrm{7b}+\mathrm{8c}=\mathrm{4} \\ $$$$\mathrm{a}+\mathrm{8c}=\mathrm{4}+\mathrm{7b} \\ $$$$\left(\mathrm{a}+\mathrm{8c}\right)^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{7b}\right)^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{16ac}+\mathrm{64c}^{\mathrm{2}} =\mathrm{16}+\mathrm{56b}+\mathrm{49b}^{\mathrm{2}} ...\left(\mathrm{1}\right) \\ $$$$\: \\ $$$$\mathrm{8a}+\mathrm{4b}−\mathrm{c}=\mathrm{7} \\ $$$$\mathrm{8a}−\mathrm{c}=\mathrm{7}−\mathrm{4b} \\ $$$$\left(\mathrm{8a}−\mathrm{c}\right)^{\mathrm{2}} =\left(\mathrm{7}−\mathrm{4b}\right)^{\mathrm{2}} \\ $$$$\mathrm{64a}^{\mathrm{2}} −\mathrm{16ac}+\mathrm{c}^{\mathrm{2}} =\mathrm{49}−\mathrm{56b}+\mathrm{16b}^{\mathrm{2}} ...\left(\mathrm{2}\right) \\ $$$$\: \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{16ac}+\mathrm{64c}^{\mathrm{2}} =\mathrm{16}+\mathrm{56b}+\mathrm{49b}^{\mathrm{2}} \\ $$$$\mathrm{64a}^{\mathrm{2}} −\mathrm{16ac}+\mathrm{c}^{\mathrm{2}} =\mathrm{49}−\mathrm{56b}+\mathrm{16b}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−+ \\ $$$$\mathrm{65a}^{\mathrm{2}} +\mathrm{65c}^{\mathrm{2}} =\mathrm{65}+\mathrm{65b}^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{1}+\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{60}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)=\mathrm{60} \\ $$

Commented by Joel575 last updated on 23/Dec/16

makasih kk terbaik

$${makasih}\:{kk} \\ $$$${terbaik} \\ $$

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